6q2-24q=3

Solution for 6q2-24q=3 equation:

6q^2-24q=3

We move all terms to the left:

6q^2-24q-(3)=0

a = 6; b = -24; c = -3;
Δ = b2-4ac
Δ = -242-4·6·(-3)
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-18\sqrt{2}}{2*6}=\frac{24-18\sqrt{2}}{12}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+18\sqrt{2}}{2*6}=\frac{24+18\sqrt{2}}{12}
$