q^2+4q=8

Solution for q^2+4q=8 equation:

q^2+4q=8

We move all terms to the left:

q^2+4q-(8)=0

a = 1; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·1·(-8)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*1}=\frac{-4-4\sqrt{3}}{2}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*1}=\frac{-4+4\sqrt{3}}{2}
$