4x^2-20x+16=-8

Solution for 4x^2-20x+16=-8 equation:

4x^2-20x+16=-8

We move all terms to the left:

4x^2-20x+16-(-8)=0

We add all the numbers together, and all the variables

4x^2-20x+24=0

a = 4; b = -20; c = +24;
Δ = b2-4ac
Δ = -202-4·4·24
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4}{2*4}=\frac{16}{8}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4}{2*4}=\frac{24}{8}
=3
$