(x-4)=x^2-8x+16

Solution for (x-4)=x^2-8x+16 equation:

(x-4)=x^2-8x+16

We move all terms to the left:

(x-4)-(x^2-8x+16)=0

We get rid of parentheses

-x^2+x+8x-4-16=0

We add all the numbers together, and all the variables

-1x^2+9x-20=0

a = -1; b = 9; c = -20;
Δ = b2-4ac
Δ = 92-4·(-1)·(-20)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*-1}=\frac{-10}{-2}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*-1}=\frac{-8}{-2}
=+4
$