-2y^2+8y+16=0

Solution for -2y^2+8y+16=0 equation:

-2y^2+8y+16=0

a = -2; b = 8; c = +16;
Δ = b2-4ac
Δ = 82-4·(-2)·16
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{3}}{2*-2}=\frac{-8-8\sqrt{3}}{-4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{3}}{2*-2}=\frac{-8+8\sqrt{3}}{-4}
$