4q^2+3q=3q^2-3q+16

Solution for 4q^2+3q=3q^2-3q+16 equation:

4q^2+3q=3q^2-3q+16

We move all terms to the left:

4q^2+3q-(3q^2-3q+16)=0

We get rid of parentheses

4q^2-3q^2+3q+3q-16=0

We add all the numbers together, and all the variables

q^2+6q-16=0

a = 1; b = 6; c = -16;
Δ = b2-4ac
Δ = 62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*1}=\frac{-16}{2}
=-8
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*1}=\frac{4}{2}
=2
$