3h^2+12h-112=0

Solution for 3h^2+12h-112=0 equation:

3h^2+12h-112=0

a = 3; b = 12; c = -112;
Δ = b2-4ac
Δ = 122-4·3·(-112)
Δ = 1488
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:

$\sqrt{\Delta}=\sqrt{1488}=\sqrt{16*93}=\sqrt{16}*\sqrt{93}=4\sqrt{93}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{93}}{2*3}=\frac{-12-4\sqrt{93}}{6}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{93}}{2*3}=\frac{-12+4\sqrt{93}}{6}
$