30=3q^2-16q+24

Solution for 30=3q^2-16q+24 equation:

30=3q^2-16q+24

We move all terms to the left:

30-(3q^2-16q+24)=0

We get rid of parentheses

-3q^2+16q-24+30=0

We add all the numbers together, and all the variables

-3q^2+16q+6=0

a = -3; b = 16; c = +6;
Δ = b2-4ac
Δ = 162-4·(-3)·6
Δ = 328
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:

$\sqrt{\Delta}=\sqrt{328}=\sqrt{4*82}=\sqrt{4}*\sqrt{82}=2\sqrt{82}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{82}}{2*-3}=\frac{-16-2\sqrt{82}}{-6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{82}}{2*-3}=\frac{-16+2\sqrt{82}}{-6}
$