4q^2=3q=3q^2-4q+18

Solution for 4q^2=3q=3q^2-4q+18 equation:

4q^2=3q=3q^2-4q+18

We move all terms to the left:

4q^2-(3q)=0

a = 4; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·4·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*4}=\frac{0}{8}
=0
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*4}=\frac{6}{8}
=3/4
$