-2b^2+4b+16=0

Solution for -2b^2+4b+16=0 equation:

-2b^2+4b+16=0

a = -2; b = 4; c = +16;
Δ = b2-4ac
Δ = 42-4·(-2)·16
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*-2}=\frac{-16}{-4}
=+4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*-2}=\frac{8}{-4}
=-2
$