4q(q+4)=-12

Solution for 4q(q+4)=-12 equation:

4q(q+4)=-12

We move all terms to the left:

4q(q+4)-(-12)=0

We add all the numbers together, and all the variables

4q(q+4)+12=0

We multiply parentheses

4q^2+16q+12=0

a = 4; b = 16; c = +12;
Δ = b2-4ac
Δ = 162-4·4·12
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*4}=\frac{-24}{8}
=-3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*4}=\frac{-8}{8}
=-1
$