8q^2-3q=7q^2-6q+18

Solution for 8q^2-3q=7q^2-6q+18 equation:

8q^2-3q=7q^2-6q+18

We move all terms to the left:

8q^2-3q-(7q^2-6q+18)=0

We get rid of parentheses

8q^2-7q^2-3q+6q-18=0

We add all the numbers together, and all the variables

q^2+3q-18=0

a = 1; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·1·(-18)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*1}=\frac{-12}{2}
=-6
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*1}=\frac{6}{2}
=3
$