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1

Solved example of precalculus

$x^2+6x+8=0$
2

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where in this case $a=1$, $b=6$ and $c=8$. Then substitute the values of the coefficients of the equation in the quadratic formula:<ul><li>$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$</li></ul>

$x=\frac{-6\pm \sqrt{6^2-4\cdot 8}}{2}$
3

Calculate the power $6^2$

$x=\frac{-6\pm \sqrt{36-4\cdot 8}}{2}$
4

Multiply $-4$ times $8$

$x=\frac{-6\pm \sqrt{36-32}}{2}$
5

Add the values $36$ and $-32$

$x=\frac{-6\pm \sqrt{4}}{2}$
6

The square root of $4$ is $2$

$x=\frac{-6\pm 2}{2}$
7

To obtain the two solutions, divide the equation in two equations, one when $\pm$ is positive ($+$), and another when $\pm$ is negative ($-$)

$x=\frac{-6+2}{2},\:x=\frac{-6-2}{2}$
8

Add the values $-6$ and $-2$

$x=\frac{-4}{2},\:x=\frac{-8}{2}$
9

Solve the equation ($1$)

$x=\frac{-4}{2}$
10

Solve the equation ($2$)

$x=\frac{-8}{2}$
11

The $2$ solutions of the equation are

$x=-2,\:x=-4$

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