-3v^2+33v+36=0

Solution for -3v^2+33v+36=0 equation:

-3v^2+33v+36=0

a = -3; b = 33; c = +36;
Δ = b2-4ac
Δ = 332-4·(-3)·36
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-39}{2*-3}=\frac{-72}{-6}
=+12
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+39}{2*-3}=\frac{6}{-6}
=-1
$