120=r^2+16r

Solution for 120=r^2+16r equation:

120=r^2+16r

We move all terms to the left:

120-(r^2+16r)=0

We get rid of parentheses

-r^2-16r+120=0

We add all the numbers together, and all the variables

-1r^2-16r+120=0

a = -1; b = -16; c = +120;
Δ = b2-4ac
Δ = -162-4·(-1)·120
Δ = 736
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:

$\sqrt{\Delta}=\sqrt{736}=\sqrt{16*46}=\sqrt{16}*\sqrt{46}=4\sqrt{46}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{46}}{2*-1}=\frac{16-4\sqrt{46}}{-2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{46}}{2*-1}=\frac{16+4\sqrt{46}}{-2}
$