r^2-24=-20r

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

r^2-24=-20r

We move all terms to the left:

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

We get rid of parentheses

r^2+20r-24=0

a = 1; b = 20; c = -24;
Δ = b2-4ac
Δ = 202-4·1·(-24)
Δ = 496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{31}}{2*1}=\frac{-20-4\sqrt{31}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{31}}{2*1}=\frac{-20+4\sqrt{31}}{2}
$