0=-0.04x^2+2x+20

Solution for 0=-0.04x^2+2x+20 equation:

0=-0.04x^2+2x+20

We move all terms to the left:

0-(-0.04x^2+2x+20)=0

We add all the numbers together, and all the variables

-(-0.04x^2+2x+20)=0

We get rid of parentheses

0.04x^2-2x-20=0

a = 0.04; b = -2; c = -20;
Δ = b2-4ac
Δ = -22-4·0.04·(-20)
Δ = 7.2
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{-(-2)-\sqrt{7.2}}{2*0.04}=\frac{2-\sqrt{7.2}}{0.08}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{7.2}}{2*0.04}=\frac{2+\sqrt{7.2}}{0.08}
$