4r^2+20r=-24

Solution for 4r^2+20r=-24 equation:

4r^2+20r=-24

We move all terms to the left:

4r^2+20r-(-24)=0

We add all the numbers together, and all the variables

4r^2+20r+24=0

a = 4; b = 20; c = +24;
Δ = b2-4ac
Δ = 202-4·4·24
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4}{2*4}=\frac{-24}{8}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4}{2*4}=\frac{-16}{8}
=-2
$