q^2+8q+16=-3q^2+6q+36

Solution for q^2+8q+16=-3q^2+6q+36 equation:

q^2+8q+16=-3q^2+6q+36

We move all terms to the left:

q^2+8q+16-(-3q^2+6q+36)=0

We get rid of parentheses

q^2+3q^2-6q+8q-36+16=0

We add all the numbers together, and all the variables

4q^2+2q-20=0

a = 4; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·4·(-20)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-18}{2*4}=\frac{-20}{8}
=-2+1/2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+18}{2*4}=\frac{16}{8}
=2
$