r^2=-2r+24

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

r^2=-2r+24

We move all terms to the left:

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

We get rid of parentheses

r^2+2r-24=0

a = 1; b = 2; c = -24;
Δ = b2-4ac
Δ = 22-4·1·(-24)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*1}=\frac{-12}{2}
=-6
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*1}=\frac{8}{2}
=4
$