20=-0.04x^2+28

Solution for 20=-0.04x^2+28 equation:

20=-0.04x^2+28

We move all terms to the left:

20-(-0.04x^2+28)=0

We get rid of parentheses

0.04x^2-28+20=0

We add all the numbers together, and all the variables

0.04x^2-8=0

a = 0.04; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·0.04·(-8)
Δ = 1.28
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{-(0)-\sqrt{1.28}}{2*0.04}=\frac{0-\sqrt{1.28}}{0.08}
=-\frac{\sqrt{}}{0.08}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{1.28}}{2*0.04}=\frac{0+\sqrt{1.28}}{0.08}
=\frac{\sqrt{}}{0.08}
$