4r^2-20=-16r

Solution for 4r^2-20=-16r equation:

4r^2-20=-16r

We move all terms to the left:

4r^2-20-(-16r)=0

We get rid of parentheses

4r^2+16r-20=0

a = 4; b = 16; c = -20;
Δ = b2-4ac
Δ = 162-4·4·(-20)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-24}{2*4}=\frac{-40}{8}
=-5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+24}{2*4}=\frac{8}{8}
=1
$