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}
$