104r+36-12r^2=0

Solution for 104r+36-12r^2=0 equation:

104r+36-12r^2=0

a = -12; b = 104; c = +36;
Δ = b2-4ac
Δ = 1042-4·(-12)·36
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(104)-112}{2*-12}=\frac{-216}{-24}
=+9
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(104)+112}{2*-12}=\frac{8}{-24}
=-1/3
$