x^2-12x+4=20

Solution for x^2-12x+4=20 equation:

x^2-12x+4=20

We move all terms to the left:

x^2-12x+4-(20)=0

We add all the numbers together, and all the variables

x^2-12x-16=0

a = 1; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·1·(-16)
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{13}}{2*1}=\frac{12-4\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{13}}{2*1}=\frac{12+4\sqrt{13}}{2}
$