20=x^2-5x

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

20=x^2-5x

We move all terms to the left:

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

We get rid of parentheses

-x^2+5x+20=0

We add all the numbers together, and all the variables

-1x^2+5x+20=0

a = -1; b = 5; c = +20;
Δ = b2-4ac
Δ = 52-4·(-1)·20
Δ = 105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{105}}{2*-1}=\frac{-5-\sqrt{105}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{105}}{2*-1}=\frac{-5+\sqrt{105}}{-2}
$