4x^2=20x+8

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

4x^2=20x+8

We move all terms to the left:

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

We get rid of parentheses

4x^2-20x-8=0

a = 4; b = -20; c = -8;
Δ = b2-4ac
Δ = -202-4·4·(-8)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{33}}{2*4}=\frac{20-4\sqrt{33}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{33}}{2*4}=\frac{20+4\sqrt{33}}{8}
$