2x^2+96x+192=0

Solution for 2x^2+96x+192=0 equation:

2x^2+96x+192=0

a = 2; b = 96; c = +192;
Δ = b2-4ac
Δ = 962-4·2·192
Δ = 7680
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{7680}=\sqrt{256*30}=\sqrt{256}*\sqrt{30}=16\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{30}}{2*2}=\frac{-96-16\sqrt{30}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{30}}{2*2}=\frac{-96+16\sqrt{30}}{4}
$