r^2+10r=24

Solution for r^2+10r=24 equation:

r^2+10r=24

We move all terms to the left:

r^2+10r-(24)=0

a = 1; b = 10; c = -24;
Δ = b2-4ac
Δ = 102-4·1·(-24)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*1}=\frac{-24}{2}
=-12
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*1}=\frac{4}{2}
=2
$