4h^2+8h-16=0

Solution for 4h^2+8h-16=0 equation:

4h^2+8h-16=0

a = 4; b = 8; c = -16;
Δ = b2-4ac
Δ = 82-4·4·(-16)
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{5}}{2*4}=\frac{-8-8\sqrt{5}}{8}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{5}}{2*4}=\frac{-8+8\sqrt{5}}{8}
$