20=5r^2

Solution for 20=5r^2 equation:

20=5r^2

We move all terms to the left:

20-(5r^2)=0

a = -5; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-5)·20
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*-5}=\frac{-20}{-10}
=+2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*-5}=\frac{20}{-10}
=-2
$