-2y^2-16y-3=0

Solution for -2y^2-16y-3=0 equation:

-2y^2-16y-3=0

a = -2; b = -16; c = -3;
Δ = b2-4ac
Δ = -162-4·(-2)·(-3)
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{58}}{2*-2}=\frac{16-2\sqrt{58}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{58}}{2*-2}=\frac{16+2\sqrt{58}}{-4}
$