3q^2+3q-54=0

Solution for 3q^2+3q-54=0 equation:

3q^2+3q-54=0

a = 3; b = 3; c = -54;
Δ = b2-4ac
Δ = 32-4·3·(-54)
Δ = 657
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:

$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{73}}{2*3}=\frac{-3-3\sqrt{73}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{73}}{2*3}=\frac{-3+3\sqrt{73}}{6}
$