20=5x^2-5x

Solution for 20=5x^2-5x equation:

20=5x^2-5x

We move all terms to the left:

20-(5x^2-5x)=0

We get rid of parentheses

-5x^2+5x+20=0

a = -5; b = 5; c = +20;
Δ = b2-4ac
Δ = 52-4·(-5)·20
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{17}}{2*-5}=\frac{-5-5\sqrt{17}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{17}}{2*-5}=\frac{-5+5\sqrt{17}}{-10}
$