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}
$