4=-20x^2+400x

Solution for 4=-20x^2+400x equation:

4=-20x^2+400x

We move all terms to the left:

4-(-20x^2+400x)=0

We get rid of parentheses

20x^2-400x+4=0

a = 20; b = -400; c = +4;
Δ = b2-4ac
Δ = -4002-4·20·4
Δ = 159680
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:

$\sqrt{\Delta}=\sqrt{159680}=\sqrt{64*2495}=\sqrt{64}*\sqrt{2495}=8\sqrt{2495}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-400)-8\sqrt{2495}}{2*20}=\frac{400-8\sqrt{2495}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-400)+8\sqrt{2495}}{2*20}=\frac{400+8\sqrt{2495}}{40}
$