Quadratic Equation Calculator

Enter the coefficients, a\hspace{0.2em} a \hspace{0.2em}, b\hspace{0.2em} b \hspace{0.2em}, and c\hspace{0.2em} c \hspace{0.2em}

x2+\hspace{0.2em} x^2\,+ \hspace{0.2em} x+\hspace{0.2em} x\,+ \hspace{0.2em} =0\hspace{0.2em} = 0 \hspace{0.2em}

Hello there!

Please provide your input and click the calculate button
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About the Quadratic Equation Calculator

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).

Usage Guide

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i. Valid Inputs

Each of the three inputs can be any real number (with one exception, mentioned below). Here are a few examples.

  • Whole numbers or decimals → 2\hspace{0.2em} 2 \hspace{0.2em}, 4.25\hspace{0.2em} -4.25 \hspace{0.2em}, 0\hspace{0.2em} 0 \hspace{0.2em}, 0.33\hspace{0.2em} 0.33 \hspace{0.2em}
  • Fractions → 2/3\hspace{0.2em} 2/3 \hspace{0.2em}, 1/5\hspace{0.2em} -1/5 \hspace{0.2em}
  • Mixed numbers → 51/4\hspace{0.2em} 5 \hspace{0.5em} 1/4 \hspace{0.2em}

NOTE — The first input, the coefficient of x2\hspace{0.2em} x^2 \hspace{0.2em}, must be a non-zero real number. (If a=0\hspace{0.2em} a = 0 \hspace{0.2em}, the second-degree term would vanish and it won't be a quadratic equation.)

ii. Example

If you would like to see an example of the calculator's working, just click the "example" button.

iii. Solutions

As mentioned earlier, the calculator won't just tell you the answer but also the steps you can follow to do the calculation yourself. The "show/hide solution" button would be available to you after the calculator has processed your input.

iv. Share

We would love to see you share our calculators with your family, friends, or anyone else who might find it useful.

By checking the "include calculation" checkbox, you can share your calculation as well.

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 ax2+bx+c=0,a0\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.

  • 5x2x+7=0\hspace{0.2em} 5x^2 - x + 7 = 0 \hspace{0.2em}
  • x2+6x=11\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, a\hspace{0.2em} a \hspace{0.2em}, cannot be 0\hspace{0.2em} 0 \hspace{0.2em} because then the second–degree term (x2\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 x\hspace{0.2em} x \hspace{0.2em} for which the equation becomes true. And those values of x\hspace{0.2em} x \hspace{0.2em} are the solutions of the equation.

As an example, for the quadratic equation x22x3=0\hspace{0.2em} x^2 - 2x - 3 = 0 \hspace{0.2em}, the solutions would be 1\hspace{0.2em} -1 \hspace{0.2em} and 3\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.

Solve the equation x2x=2\hspace{0.2em} x^2 - x = 2 \hspace{0.2em}.


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

In this example, we have

x2x=2x^2 - x = 2

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

x2x2=0x^2 - x - 2 = 0

Step 2.  Factor the quadratic trinomial.

I am assuming you are comfortable with factoring a trinomial.

x2x2=0(x+1)(x2)=0\begin{align*} &x^2 - x - 2 = 0 \\[1em] &(x + 1)(x - 2) = 0 \end{align*}

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


x+1=0x=1\begin{align*} x + 1 &= 0 \\[1em] x &= -1 \end{align*}
x2=0x=2\begin{align*} x - 2 &= 0 \\[1em] x &= 2 \end{align*}

And that's it. Our equation has two real and distinct solutions (1,2)\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.

Using the Quadratic Formula

Consider the following quadratic equation.

ax2+bx+c=0ax^2 + bx + c = 0

According to the quadratic formula, the solutions (x1,x2)\hspace{0.2em} (x_1, x_2) \hspace{0.2em} to this quadratic equation would be —

x1,x2=b±b24ac2ax_1, x_2 \hspace{0.25em} = \hspace{0.25em} \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Here's an example.

Solve the quadratic equation 2x26x+1=0\hspace{0.2em} 2x^2 - 6x + 1 = 0 \hspace{0.2em} using the quadratic formula.


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

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

a=2,b=6,c=1a \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.

x1,x2=b±b24ac2a=(6)±(6)242122=6±284=3±72\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 —

x1=3+72,x2=372x_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.

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