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A quadratic equation is an equation formed by putting a quadratic polynomial equal to zero.

## Standard Form

The standard form of the quadratic equation is a polynomial of the second degree, written as $ax²+bx+c=0$, where $x$ is unknown and $a, b, c$ are real numbers with $a \ne 0$. The equation is not quadratic if $a=0$, because the term $ax²$ is reduced to zero leaving only $bx+c=0$, which is the equation of a line.

The easiest way to solve a quadratic equation is probably the quadratic formula. The quadratic formula can be derived by completing the square on the quadratic equation $ax²+bx+c=0$. The quadratic formula is:

$\\left(\textrm\left\{x=\right\}\\right)$$$\frac{-\textrm{b}\pm\sqrt{\textrm{b}^2-\textrm{4ac}}}{\textrm{2a}}$$$\left(a \ne 0\right)$

Using the quadratic formula to solve a quadratic equation is straight-forward, as shown by the following example:

Given the equation $x²+10x+21=0$, we get:

$\\left(\textrm\left\{x=\right\}\frac\left\{-10\pm\sqrt\left\{10^2-4\left(1\right)\left(21\right)\right\}\right\}\left\{2\left(1\right)\right\}\\right)$

$\\left(\,\,\textrm\left\{=\right\}\frac\left\{-10\pm\sqrt\left\{100-84\right\}\right\}\left\{2\right\}\\right)$

$\\left(\,\,\textrm\left\{=\right\}\frac\left\{-10\pm\sqrt\left\{16\right\}\right\}\left\{2\right\}\\right)$

$\\left(\,\,\textrm\left\{=\right\}\frac\left\{-10\pm 4\right\}\left\{2\right\}\\right)$

$\\left(\,\,\textrm\left\{=\right\}\\left\{\frac\left\{-10-4\right\}\left\{2\right\}, \frac\left\{-10+4\right\}\left\{2\right\}\\right\}\\right)$

$\\left(\,\,\textrm\left\{=\right\}\\left\{\frac\left\{-14\right\}\left\{2\right\}, \frac\left\{-6\right\}\left\{2\right\}\\right\}\\right)$

$\\left(\,\,\textrm\left\{=\right\}\\left\{\\right)-7, -3\\left(\\right\}\\right)$

### Solution by Factoring

Solving quadratic equations can be done using various methods. One way to solve a quadratic equation is by factoring. It is important to keep in mind, however, that not every quadratic equation can be factored. Factoring a trinomial expression is part of another topic (please refer to Factoring Polynomials for a detailed step-by-step explanation.)

Once the quadratic equation is factored, the roots of the equation can be easily found. For example, the quadratic expression $x²+7x+12$ factors to $\left(x+4\right)\left(x+3\right)$. Since assigning a value of either $-3 or -4$ to $x$ satisifies the equation $x²+7x+12=0$, the solution set is $x=\left\{-4,-3\right\}$.

A slightly more complicated example is one such as $6x²+6x-12$. After factoring, it becomes $\left(2x+4\right)\left(3x-3\right)=0$. We then need to solve each of the roots algebraically:

The solution set is therefore $x=\left\{-2, 1\right\}$.

### Completing the Square

Since it's not possible to factor every quadratic expression, we often solve quadratic equations by the method known as completing the square. In order to solve a quadratic equation by completing the square, we perform the following steps:

1) Write the equation on the standard form $ax²+bx+c=0$.
2) Divide both sides of the equation by the coefficient of $x²$ if not $1$ (i.e, divide by $a$.)
3) Subtract the constant term from both sides (i.e, subtract $c$.)
4) Divide the coefficient of $x$ by $2$, square the result, then add the resulting value to both sides (i.e, add $\left(½b\right)²$ or $¼b²$.)
5) Factor the left side of the equation.
6) Apply the square root property $±\surd \dots$
7) Check the results if required. (i.e, plug the results into the equation to test it.)
8) Write the solution set.

For example, given the quadratic equation $3x²-10x-2$, we follow the steps given above:

1) Write the equation in standard form:
$3x²-10x-2=0$

2) Divide both sides of the equation by the coefficient of $x²$, which is $3$:
$x²-\\left(\frac\left\{10\right\}\left\{3\right\}\\right)x-\\left(\frac\left\{2\right\}\left\{3\right\}\\right)=0$

3) Subtract the constant term from both sides:
$x²-\\left(\frac\left\{10\right\}\left\{3\right\}\\right)x=\\left(\frac\left\{2\right\}\left\{3\right\}\\right)$

4) Divide the coefficient of $x$ by $2$, square the result, then add the resulting value to both sides:
(-$$\frac{10}{3}$$⋅$$\frac{1}{2}$$)²=-($$\frac{5}{3}$$)²=$$\frac{25}{9}$$

$x²-\\left(\frac\left\{10\right\}\left\{3\right\}\\right)+x+\\left(\frac\left\{25\right\}\left\{9\right\}\\right)=\\left(\frac\left\{2\right\}\left\{3\right\}\\right)+\\left(\frac\left\{25\right\}\left\{9\right\}\\right)=\\left(\frac\left\{31\right\}\left\{9\right\}\\right)$

5) Factor the left side of the equation:
($x-\\left(\frac\left\{5\right\}\left\{3\right\}\\right)$)²=$$\frac{31}{9}$$

6) Apply the square root property:
$x-\\left(\frac\left\{5\right\}\left\{3\right\}\\right)= ±\\left(\sqrt\left\{\frac\left\{31\right\}\left\{9\right\}\right\}\\right)$

$x-\\left(\frac\left\{5\right\}\left\{3\right\}\\right)= ±\\left(\frac\left\{\sqrt\left\{31\right\}\right\}\left\{3\right\}\\right)$

$x=\\left(\frac\left\{5\right\}\left\{3\right\}\\right)±\\left(\frac\left\{\sqrt\left\{31\right\}\right\}\left\{3\right\}\\right)$

7) Check the results if required:
$3$($$\frac{5}{3}+\frac{\sqrt{31}}{3}$$)²$-10$($$\frac{5}{3}+\frac{\sqrt{31}}{3}$$)$-2=0$

$3$($\\left(\frac\left\{5\right\}\left\{3\right\}-\frac\left\{\sqrt\left\{31\right\}\right\}\left\{3\right\}\\right)$)²$-10$($$\frac{5}{3}-\frac{\sqrt{31}}{3}$$)$-2=0$

8) Write the solution set:
$x =${$$\frac{5-\sqrt{31}}{3}$$, $$\frac{5+\sqrt{31}}{3}$$}  (or)  $x =${$$\frac{5±\sqrt{31}}{3}$$}

Since completing the square is more complex than factoring or using the quadratic formula, it's always a good idea to try to use those methods before using this one (that is, unless you're doing homework and the assignment is to solve the problem by completing the square!)