2y^2+14y-192=0

Solution for 2y^2+14y-192=0 equation:

2y^2+14y-192=0

a = 2; b = 14; c = -192;
Δ = b2-4ac
Δ = 142-4·2·(-192)
Δ = 1732
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1732}=\sqrt{4*433}=\sqrt{4}*\sqrt{433}=2\sqrt{433}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{433}}{2*2}=\frac{-14-2\sqrt{433}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{433}}{2*2}=\frac{-14+2\sqrt{433}}{4}
$