4x2+20=20x

Solution for 4x2+20=20x equation:

4x^2+20=20x

We move all terms to the left:

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

a = 4; b = -20; c = +20;
Δ = b2-4ac
Δ = -202-4·4·20
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{5}}{2*4}=\frac{20-4\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{5}}{2*4}=\frac{20+4\sqrt{5}}{8}
$