q^2+8q+16=-2q^2+9q+436

Solution for q^2+8q+16=-2q^2+9q+436 equation:

q^2+8q+16=-2q^2+9q+436

We move all terms to the left:

q^2+8q+16-(-2q^2+9q+436)=0

We get rid of parentheses

q^2+2q^2-9q+8q-436+16=0

We add all the numbers together, and all the variables

3q^2-1q-420=0

a = 3; b = -1; c = -420;
Δ = b2-4ac
Δ = -12-4·3·(-420)
Δ = 5041
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5041}=71$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-71}{2*3}=\frac{-70}{6}
=-11+2/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+71}{2*3}=\frac{72}{6}
=12
$