Question

Am I correct? I just want to know if I got it correct.

Answer 1

We want to solve the following equation

$x^2-12x+36=80$

we begin by subtracting 80 on both sides, so we get

$x^2-12x+36-80=x^2\text{ -12x -44}=0$

So we have the equivalent problem of solving the equation

$x^2\text{ -12x-44=0}$

Recall that having an equation of the form

$ax^2+bx+c=0^{}$

the solutions are given by the equation

$x=\frac{\text{ -b }\pm\sqrt[]{b^2\text{ -4ac}}}{2a}$

In our case we have a=1, b=-12 and c= -44 so the solutions are

$x=\frac{\text{ -(-12)}\pm\sqrt[]{(-12)^2\text{ -4(1)( -44)}}}{2(1)}=\frac{12\pm\sqrt[]{320}}{2}=\frac{12\pm8\sqrt[]{5}}{2}=6\pm4\sqrt[\square]{5}$

so the solutions to the original problem are

$x=6+4\sqrt[]{5}$

and

$x=6\text{ -4}\sqrt[]{5}$