3q2=2q+8

Solution for 3q2=2q+8 equation:

3q^2=2q+8

We move all terms to the left:

3q^2-(2q+8)=0

We get rid of parentheses

3q^2-2q-8=0

a = 3; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·3·(-8)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*3}=\frac{-8}{6}
=-1+1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*3}=\frac{12}{6}
=2
$