8q^2-3q=7q^2-9q=+16

Solution for 8q^2-3q=7q^2-9q=+16 equation:

8q^2-3q=7q^2-9q=+16

We move all terms to the left:

8q^2-3q-(7q^2-9q)=0

We get rid of parentheses

8q^2-7q^2-3q+9q=0

We add all the numbers together, and all the variables

q^2+6q=0

a = 1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*1}=\frac{-12}{2}
=-6
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*1}=\frac{0}{2}
=0
$