100=20x+4x^2

Solution for 100=20x+4x^2 equation:

100=20x+4x^2

We move all terms to the left:

100-(20x+4x^2)=0

We get rid of parentheses

-4x^2-20x+100=0

a = -4; b = -20; c = +100;
Δ = b2-4ac
Δ = -202-4·(-4)·100
Δ = 2000
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:

$\sqrt{\Delta}=\sqrt{2000}=\sqrt{400*5}=\sqrt{400}*\sqrt{5}=20\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{5}}{2*-4}=\frac{20-20\sqrt{5}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{5}}{2*-4}=\frac{20+20\sqrt{5}}{-8}
$