q^2+14=-3q^2-6q+87

Solution for q^2+14=-3q^2-6q+87 equation:

q^2+14=-3q^2-6q+87

We move all terms to the left:

q^2+14-(-3q^2-6q+87)=0

We get rid of parentheses

q^2+3q^2+6q-87+14=0

We add all the numbers together, and all the variables

4q^2+6q-73=0

a = 4; b = 6; c = -73;
Δ = b2-4ac
Δ = 62-4·4·(-73)
Δ = 1204
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1204}=\sqrt{4*301}=\sqrt{4}*\sqrt{301}=2\sqrt{301}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{301}}{2*4}=\frac{-6-2\sqrt{301}}{8}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{301}}{2*4}=\frac{-6+2\sqrt{301}}{8}
$