-3q^2-15q+12=0

Solution for -3q^2-15q+12=0 equation:

-3q^2-15q+12=0

a = -3; b = -15; c = +12;
Δ = b2-4ac
Δ = -152-4·(-3)·12
Δ = 369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{41}}{2*-3}=\frac{15-3\sqrt{41}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{41}}{2*-3}=\frac{15+3\sqrt{41}}{-6}
$