# How to solve quadratic equations

Canonical: https://duckyhelper.com/learn/algebra-1/solving-quadratics/
Updated: 2026-10-01

To solve a quadratic equation, first write it as \(ax^2 + bx + c = 0\). If it factors, set each factor equal to zero. If there is no x term, isolate \(x^2\) and take the square root of both sides, keeping both the positive and negative root. The quadratic formula works every time. The discriminant, \(b^2 - 4ac\), tells you whether there are two, one or no real solutions.

## The key idea

Factoring works because of the **zero product property**: if two numbers multiply to 0, at least one of them is 0. That is why the equation must equal zero before you factor.

$$
(x + 7)(x - 4) = 0 \;\Longrightarrow\; x + 7 = 0 \;\text{ or }\; x - 4 = 0
$$

When factoring is hard or impossible, the quadratic formula solves any \(ax^2 + bx + c = 0\):

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

**Which method?**

| The equation looks like | Use | Example |
| --- | --- | --- |
| no x term, or a squared group | square roots | \(3(x - 2)^2 = 75\) |
| easy whole-number factors | factoring | \(x^2 + 3x - 28 = 0\) |
| anything else | quadratic formula | \(2x^2 - 4x - 3 = 0\) |

**The discriminant b^2 - 4ac**

| If it is | Real solutions | The graph \(y = ax^2 + bx + c\) |
| --- | --- | --- |
| positive | two | crosses the x-axis twice |
| zero | one (a double root) | touches the x-axis once |
| negative | none | never reaches the x-axis |

## Worked examples

**Example 1: factoring**

Problem: Solve \(x^2 + 3x = 28\).

1. Get zero on one side first. Subtract 28.

   $$
   x^2 + 3x - 28 = 0
   $$
2. Find two numbers that multiply to \(-28\) and add to 3: they are 7 and \(-4\).

   $$
   (x + 7)(x - 4) = 0
   $$
3. Set each factor equal to zero.

   $$
   x + 7 = 0 \quad\text{or}\quad x - 4 = 0
   $$
4. Solve each one.

   $$
   x = -7 \quad\text{or}\quad x = 4
   $$

Answer: \(x = -7\) or \(x = 4\)

**Example 2: square roots**

Problem: Solve \(3(x - 2)^2 = 75\).

1. Divide both sides by 3 to get the squared group alone.

   $$
   (x - 2)^2 = 25
   $$
2. Take the square root of both sides. Both 5 and \(-5\) square to 25, so write both.

   $$
   x - 2 = 5 \quad\text{or}\quad x - 2 = -5
   $$
3. Add 2 in each case.

   $$
   x = 7 \quad\text{or}\quad x = -3
   $$

Answer: \(x = 7\) or \(x = -3\)

**Example 3: the quadratic formula**

Problem: Solve \(2x^2 - 4x - 3 = 0\).

1. Here \(a = 2\), \(b = -4\), \(c = -3\). Work out the discriminant first. Note that \((-4)^2\) is positive 16.

   $$
   (-4)^2 - 4(2)(-3) = 16 + 24 = 40
   $$
2. Put everything into the formula. \(-b\) is \(-(-4) = 4\), and \(2a = 4\).

   $$
   x = \frac{4 + \sqrt{40}}{4} \quad\text{or}\quad x = \frac{4 - \sqrt{40}}{4}
   $$
3. Simplify: \(\sqrt{40} = 2\sqrt{10}\), then divide every term on top by the 4 (that is, divide top and bottom by 2).

   $$
   x = \frac{2 + \sqrt{10}}{2} \quad\text{or}\quad x = \frac{2 - \sqrt{10}}{2}
   $$

Answer: \(x = \frac{2 \pm \sqrt{10}}{2}\), about 2.58 and \(-0.58\)

**Example 4 (test-hard): a word problem with one answer that makes sense**

Problem: A ball is thrown upward from a 4-foot platform. Its height in feet after t seconds is \(h = -16t^2 + 48t + 4\). When does it hit the ground? Round to the nearest hundredth of a second.

1. The ground is height 0.

   $$
   -16t^2 + 48t + 4 = 0
   $$
2. Divide every term by \(-4\) to make the numbers smaller and a positive.

   $$
   4t^2 - 12t - 1 = 0
   $$
3. It does not factor, so use the formula with \(a = 4\), \(b = -12\), \(c = -1\). Discriminant:

   $$
   (-12)^2 - 4(4)(-1) = 144 + 16 = 160
   $$
4. Formula: \(t = \frac{12 \pm \sqrt{160}}{8}\). Since \(\sqrt{160} = 4\sqrt{10}\), divide top and bottom by 4.

   $$
   t = \frac{3 + \sqrt{10}}{2} \quad\text{or}\quad t = \frac{3 - \sqrt{10}}{2}
   $$
5. The second value is about \(-0.08\). Negative time is before the throw, so it does not fit the story. The first is about 3.08.

Answer: About 3.08 seconds, which is \(t = \frac{3 + \sqrt{10}}{2}\).

## Common mistakes

- **Factoring before setting the equation to zero.** \(x(x + 3) = 28\) does not mean \(x = 28\). The zero product property only works with 0 on one side.
- **Forgetting the negative root.** \((x - 2)^2 = 25\) gives \(x - 2 = 5\) **or** \(x - 2 = -5\).
- **Dividing both sides by x.** From \(x^2 = 5x\), dividing by x loses the answer \(x = 0\). Fix: move everything to one side and factor: \(x(x - 5) = 0\).
- **Sign errors with a negative b.** If \(b = -4\), then \(-b = 4\) and \(b^2 = 16\). Fix: write \(-(-4)\) and \((-4)^2\) with parentheses.
- **Dividing only part of the top by 2a.** The fraction bar covers the whole \(-b \pm \sqrt{b^2 - 4ac}\).

## Quick methods

> **Tip: Check the discriminant before you start**
>
> \(b^2 - 4ac\) takes a few seconds. If it is negative, there are no real solutions and you are done. If it is a perfect square, the equation factors with whole numbers. If it is positive but not a perfect square, go straight to the formula.

> **Note: Check with a graph**
>
> Graph \(y = ax^2 + bx + c\) in a graphing calculator like Desmos. The x-intercepts are the solutions. For Example 3, the graph crosses at about 2.58 and \(-0.58\), which confirms the exact answers.

## Practice

**5 practice questions**

1. Solve \(x^2 - 5x - 14 = 0\).
   A. \(x = 7\) or \(x = -2\)
   B. \(x = -7\) or \(x = 2\)
   C. \(x = 14\) or \(x = -1\)
   D. \(x = 5\) or \(x = -14\)

   Answer: \(x = 7\) or \(x = -2\). Factor: \((x - 7)(x + 2) = 0\), since \(-7 \cdot 2 = -14\) and \(-7 + 2 = -5\). Then \(x = 7\) or \(x = -2\). The choice \(-7\) or 2 takes the numbers from the factors without changing their signs.

2. Solve \(x^2 = 6x\).
   A. \(x = 6\)
   B. \(x = 0\) or \(x = 6\)
   C. \(x = 0\) or \(x = -6\)
   D. \(x = \sqrt{6}\)

   Answer: \(x = 0\) or \(x = 6\). Move everything to one side: \(x^2 - 6x = 0\), so \(x(x - 6) = 0\). Both 0 and 6 work. Dividing both sides by x gives only 6 and loses the 0.

3. Solve \((x + 1)^2 = 49\).
   A. \(x = 6\)
   B. \(x = 6\) or \(x = -8\)
   C. \(x = 48\)
   D. \(x = 8\) or \(x = -6\)

   Answer: \(x = 6\) or \(x = -8\). Square root both sides: \(x + 1 = 7\) or \(x + 1 = -7\). So \(x = 6\) or \(x = -8\). Just 6 misses the negative root, and 8 or \(-6\) adds 1 instead of subtracting it.

4. How many real solutions does \(3x^2 - 2x + 5 = 0\) have?
   A. 0
   B. 1
   C. 2
   D. infinitely many

   Answer: 0. The discriminant is \((-2)^2 - 4(3)(5) = 4 - 60 = -56\). It is negative, so there are no real solutions: the parabola never reaches the x-axis.

5. Solve \(x^2 + 4x - 1 = 0\).
   A. \(x = -2 \pm \sqrt{5}\)
   B. \(x = 2 \pm \sqrt{5}\)
   C. \(x = -2 \pm 2\sqrt{5}\)
   D. \(x = -4 \pm \sqrt{5}\)

   Answer: \(x = -2 \pm \sqrt{5}\). Formula: \(x = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}\). The choice \(-2 \pm 2\sqrt{5}\) divides the \(-4\) by 2 but not the root. \(2 \pm \sqrt{5}\) forgets the minus in \(-b\).

## Frequently asked questions

### Which method should I use to solve a quadratic?

If there is no x term, or you see a squared group like \((x - 2)^2\), use square roots. If the numbers factor easily, factor. Otherwise use the quadratic formula, which always works. Completing the square also always works and is how the formula is built.

### What does the discriminant tell you?

The discriminant is \(b^2 - 4ac\), the part under the square root in the formula. Positive means two real solutions, zero means exactly one, and negative means none, because you cannot take the square root of a negative number with real numbers.

### Why do quadratic equations have two answers?

The graph of a quadratic is a U-shaped parabola, and a U can cross a horizontal line in two places. Algebraically, both a number and its opposite square to the same value. Sometimes the two answers are the same (one solution), and sometimes there are none.

### What if I get a negative number under the square root?

Then the equation has no real solutions. In Algebra 1 you write "no real solution." In Algebra 2 you learn imaginary numbers, which give two complex solutions instead. Double-check that \(b^2\) was positive and the signs of a and c are right first.

## Related

- [How to factor polynomials](https://duckyhelper.com/learn/algebra-1/factoring/)
- [Completing the square](https://duckyhelper.com/learn/algebra-2/completing-the-square/)
- [Quadratics on the SAT](https://duckyhelper.com/learn/sat-math/quadratics/)
- [Algebra 1 study guides](https://duckyhelper.com/learn/algebra-1/)

## Try asking Ducky

- "Should I factor this or use the quadratic formula?"
- "I got x = 6 but the answer also has -8. Where does the second one come from?"
- "Plug this into the quadratic formula with me one step at a time."

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