2q2-8=3q

Solution for 2q2-8=3q equation:

2q^2-8=3q

We move all terms to the left:

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

a = 2; b = -3; c = -8;
Δ = b2-4ac
Δ = -32-4·2·(-8)
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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