How to Solve Quadratic Equations: 4 Methods with Step-by-Step Examples

What is a quadratic equation?

A quadratic equation is any equation you can rewrite in the form

where $a$, $b$, and $c$ are numbers and $a \ne 0$. The graph of $y = ax^2 + bx + c$ is a parabola, and the solutions of the equation — also called the roots or zeros — are exactly the $x$-values where that parabola crosses the $x$-axis. A quadratic always has 0, 1, or 2 real solutions.

There are four standard ways to solve them. The right one depends on what the equation looks like.

Method 1: Factoring

If the quadratic factors into a product of two linear pieces, factoring is fastest.

The idea. Rewrite

and use the zero-product property: if a product equals zero, at least one factor equals zero. So set each factor to zero and solve.

Example 1. Solve $x^2 - 5x + 6 = 0$.

1. Look for two numbers that multiply to $6$ and add to $-5$. Those are $-2$ and $-3$.
2. Factor: $(x - 2)(x - 3) = 0$.
3. Set each factor to zero: $x - 2 = 0$ or $x - 3 = 0$.
4. So $x = 2$ or $x = 3$.

When to use it. When $c$ has small integer factors that visibly add to $b$. If you cannot guess the factors after a few seconds, switch methods.

Method 2: The square root method

If the equation has the form $x^2 = k$ — no $x$ term — take the square root of both sides.

Example 2. Solve $3x^2 - 27 = 0$.

1. Isolate the $x^2$: $3x^2 = 27$, so $x^2 = 9$.
2. Take the square root of both sides: $x = \pm \sqrt{9} = \pm 3$.
3. So $x = 3$ or $x = -3$.

Don't forget the $\pm$ — both the positive and negative roots are valid.

When to use it. Whenever the linear coefficient $b$ is zero, or whenever the quadratic is already in the form $(x + h)^2 = k$.

Method 3: Completing the square

This method rewrites the quadratic so you can apply the square root method. It is also the derivation behind the quadratic formula.

The recipe for $x^2 + bx + c = 0$ (when $a = 1$):

1. Move $c$ to the right side: $x^2 + bx = -c$.
2. Take half of $b$, square it, and add $\left(\dfrac{b}{2}\right)^2$ to both sides.
3. The left side is now a perfect square: $\left(x + \dfrac{b}{2}\right)^2 = -c + \left(\dfrac{b}{2}\right)^2$.
4. Take the square root of both sides and solve for $x$.

Example 3. Solve $x^2 + 6x + 5 = 0$ by completing the square.

1. Rearrange: $x^2 + 6x = -5$.
2. Half of $6$ is $3$; squared is $9$. Add $9$ to both sides: $x^2 + 6x + 9 = 4$.
3. The left side is $(x + 3)^2$, so $(x + 3)^2 = 4$.
4. Square root: $x + 3 = \pm 2$, giving $x = -1$ or $x = -5$.

If $a \ne 1$, divide both sides by $a$ first, then complete the square as above.

When to use it. Useful when factoring fails and you want a clean exact answer. Completing the square is also the standard way to convert $ax^2 + bx + c$ into vertex form $a(x - h)^2 + k$, which gives you the parabola's vertex directly.

Method 4: The quadratic formula

The quadratic formula always works. For $ax^2 + bx + c = 0$ with $a \ne 0$:

The expression under the square root, $b^2 - 4ac$, is called the discriminant $\Delta$. It tells you how many real solutions the equation has before you finish the calculation:

• $\Delta > 0$: two distinct real solutions.
• $\Delta = 0$: one repeated real solution (the parabola just touches the $x$-axis).
• $\Delta < 0$: no real solutions (the parabola misses the $x$-axis).

Example 4. Solve $2x^2 + 3x - 5 = 0$.

1. Identify $a = 2$, $b = 3$, $c = -5$.
2. Compute the discriminant: $\Delta = 3^2 - 4(2)(-5) = 9 + 40 = 49$.
3. Plug into the formula: $x = \dfrac{-3 \pm \sqrt{49}}{4} = \dfrac{-3 \pm 7}{4}$.
4. So $x = \dfrac{4}{4} = 1$ or $x = \dfrac{-10}{4} = -\dfrac{5}{2}$.

When to use it. When the equation does not factor nicely and you don't want to bother completing the square. Many textbooks recommend always checking the discriminant first — if $\Delta < 0$, you know the answer is "no real solutions" and you can stop.

Choosing a method quickly

| Situation | Best method | |---|---| | Coefficients are small integers and you can factor mentally | Factoring | | No $x$ term, or already in $(x - h)^2 = k$ form | Square root | | Need vertex form, or coefficients are messy fractions | Completing the square | | Anything else, especially with irrational solutions | Quadratic formula |

If you are stuck, the quadratic formula always works — it just costs a little more arithmetic.

Common mistakes

• Forgetting $\pm$ when taking a square root. $\sqrt{9}$ gives a positive number, but $x^2 = 9$ has two solutions: $x = 3$ and $x = -3$.
• Sign errors inside the discriminant. $b^2 - 4ac$ — the $-4ac$ flips sign whenever $a$ or $c$ is negative.
• Dividing by something containing $x$. Never divide both sides by $x$ to "simplify"; you will lose the solution $x = 0$. Always move everything to one side and solve a quadratic.
• Using the formula on a non-quadratic. The quadratic formula needs $a \ne 0$. If the leading coefficient is zero, you actually have a linear equation.