2q2+4q+-96=0

Solution for 2q2+4q+-96=0 equation:

2q^2+4q+-96=0

We add all the numbers together, and all the variables

2q^2+4q=0

a = 2; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·2·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*2}=\frac{-8}{4}
=-2
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*2}=\frac{0}{4}
=0
$