x^2-8x+32=12x-4

Solution for x^2-8x+32=12x-4 equation:

x^2-8x+32=12x-4

We move all terms to the left:

x^2-8x+32-(12x-4)=0

We get rid of parentheses

x^2-8x-12x+4+32=0

We add all the numbers together, and all the variables

x^2-20x+36=0

a = 1; b = -20; c = +36;
Δ = b2-4ac
Δ = -202-4·1·36
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16}{2*1}=\frac{36}{2}
=18
$