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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -((x-8)), so:

(x-8)

We get rid of parentheses

x-8

Back to the equation:

-(x-8)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-13x+40=0

a = 1; b = -13; c = +40;
Δ = b2-4ac
Δ = -132-4·1·40
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*1}=\frac{10}{2}
=5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*1}=\frac{16}{2}
=8
$