8h^2+12h-396=0

Solution for 8h^2+12h-396=0 equation:

8h^2+12h-396=0

a = 8; b = 12; c = -396;
Δ = b2-4ac
Δ = 122-4·8·(-396)
Δ = 12816
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:

$\sqrt{\Delta}=\sqrt{12816}=\sqrt{144*89}=\sqrt{144}*\sqrt{89}=12\sqrt{89}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12\sqrt{89}}{2*8}=\frac{-12-12\sqrt{89}}{16}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12\sqrt{89}}{2*8}=\frac{-12+12\sqrt{89}}{16}
$