3q^2=28q+32

Solution for 3q^2=28q+32 equation:

3q^2=28q+32

We move all terms to the left:

3q^2-(28q+32)=0

We get rid of parentheses

3q^2-28q-32=0

a = 3; b = -28; c = -32;
Δ = b2-4ac
Δ = -282-4·3·(-32)
Δ = 1168
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:

$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{73}}{2*3}=\frac{28-4\sqrt{73}}{6}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{73}}{2*3}=\frac{28+4\sqrt{73}}{6}
$