x^2-12x+16=20

Solution for x^2-12x+16=20 equation:

x^2-12x+16=20

We move all terms to the left:

x^2-12x+16-(20)=0

We add all the numbers together, and all the variables

x^2-12x-4=0

a = 1; b = -12; c = -4;
Δ = b2-4ac
Δ = -122-4·1·(-4)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{10}}{2*1}=\frac{12-4\sqrt{10}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{10}}{2*1}=\frac{12+4\sqrt{10}}{2}
$