2h^2+8h-128=0

Solution for 2h^2+8h-128=0 equation:

2h^2+8h-128=0

a = 2; b = 8; c = -128;
Δ = b2-4ac
Δ = 82-4·2·(-128)
Δ = 1088
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:

$\sqrt{\Delta}=\sqrt{1088}=\sqrt{64*17}=\sqrt{64}*\sqrt{17}=8\sqrt{17}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{17}}{2*2}=\frac{-8-8\sqrt{17}}{4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{17}}{2*2}=\frac{-8+8\sqrt{17}}{4}
$