3q^2-4q=2q^2-6q+15

Solution for 3q^2-4q=2q^2-6q+15 equation:

3q^2-4q=2q^2-6q+15

We move all terms to the left:

3q^2-4q-(2q^2-6q+15)=0

We get rid of parentheses

3q^2-2q^2-4q+6q-15=0

We add all the numbers together, and all the variables

q^2+2q-15=0

a = 1; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·1·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*1}=\frac{-10}{2}
=-5
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*1}=\frac{6}{2}
=3
$