This quadratic equation calculator lets you calculate the roots or solutions for a quadratic equation.

The calculator will tell you not only the roots but also how to solve the quadratic equation using the quadratic formula as well as the factoring method (wherever practical).

Here's a quick overview of what quadratic equations are and how to solve them.

## Quadratic Equations

A quadratic equation is a second–degree equation with only one variable. The standard form of a quadratic equation is $\hspace{0.2em} ax^2 + bx +c = 0, \hspace{0.3em} a \neq 0 \hspace{0.2em}$.

Here are a couple of examples of quadratic equations.

- $\hspace{0.2em} 5x^2 - x + 7 = 0 \hspace{0.2em}$
- $\hspace{0.2em} x^2 + 6x = 11 \hspace{0.2em}$

The second equation is an example of a quadratic equation not in its standard form.

The leading coefficient, $\hspace{0.2em} a \hspace{0.2em}$, cannot be $\hspace{0.2em} 0 \hspace{0.2em}$ because then the second–degree term ($\hspace{0.2em} x^2$–term) would vanish and our equation would no longer be a second–degree equation.

## Solving Quadratic Equations

To solve a quadratic equation is to find the values of $\hspace{0.2em} x \hspace{0.2em}$ for which the equation becomes true. And those values of $\hspace{0.2em} x \hspace{0.2em}$ are the solutions of the equation.

As an example, for the quadratic equation $\hspace{0.2em} x^2 - 2x - 3 = 0 \hspace{0.2em}$, the solutions would be $\hspace{0.2em} -1 \hspace{0.2em}$ and $\hspace{0.2em} 3 \hspace{0.2em}$.

Now there are two popular ways of solving a quadratic equation. Let's look at them one by one.

### Factorization Method

Let me explain the factorization method using the following example.

Example

Solve the equation $\hspace{0.2em} x^2 - x = 2 \hspace{0.2em}$.

Solution

Step 1. Make sure you have every non-zero term on one side.

In this example, we have

$x^2 - x = 2$

Subtracting $\hspace{0.2em} 2 \hspace{0.2em}$ from both sides, we have

$x^2 - x - 2 = 0$

Step 2. Factor the quadratic trinomial.

I am assuming you are comfortable with factoring a trinomial.

$\begin{align*} &x^2 - x - 2 = 0 \\[1em] &(x + 1)(x - 2) = 0 \end{align*}$

Step 3. Equate each factor to $\hspace{0.2em} 0 \hspace{0.2em}$ and solve for the unknown.

So,

$\begin{align*} x + 1 &= 0 \\[1em] x &= -1 \end{align*}$

$\begin{align*} x - 2 &= 0 \\[1em] x &= 2 \end{align*}$

And that's it. Our equation has two real and distinct solutions $\hspace{0.2em} (-1, 2) \hspace{0.2em}$.

Note — While this method can be simple and quick in many cases, it's not always possible (or practical) to factor the trinomial. Thankfully, we have another method.

Consider the following quadratic equation.

$ax^2 + bx + c = 0$

According to the quadratic formula, the solutions $\hspace{0.2em} (x_1, x_2) \hspace{0.2em}$ to this quadratic equation would be —

$x_1, x_2 \hspace{0.25em} = \hspace{0.25em} \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here's an example.

Example

Solve the quadratic equation $\hspace{0.2em} 2x^2 - 6x + 1 = 0 \hspace{0.2em}$ using the quadratic formula.

Solution

To solve the equation using the quadratic formula, we need to first identify the coefficients $\hspace{0.2em} a \hspace{0.2em}$, $\hspace{0.2em} b \hspace{0.2em}$, and $\hspace{0.2em} c \hspace{0.2em}$.

So comparing the given equation with the standard equation, we get

$a \hspace{0.2em} = \hspace{0.2em} 2, \hspace{1em} b \hspace{0.2em} = \hspace{0.2em} -6, \hspace{1em} c \hspace{0.2em} = \hspace{0.2em} 1$

Next, we substitute these values into the quadratic formula.

$\begin{align*} x_1, x_2 \hspace{0.25em} &= \hspace{0.25em} \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[1.75em] &= \hspace{0.25em} \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \\[1.75em] &= \hspace{0.25em} \frac{6 \pm \sqrt{28}}{4} \\[1.75em] &= \hspace{0.25em} \frac{3 \pm \sqrt{7}}{2} \end{align*}$

Separating the two values, we get —

$x_1 \hspace{0.2em} = \hspace{0.2em} \frac{3 + \sqrt{7}}{2}, \hspace{1em} x_2 \hspace{0.2em} = \hspace{0.2em} \frac{3 - \sqrt{7}}{2}$

Equation solved.