0=140-96h+12(h^2)

Solution for 0=140-96h+12(h^2) equation:

0=140-96h+12(h^2)

We move all terms to the left:

0-(140-96h+12(h^2))=0

We add all the numbers together, and all the variables

-(140-96h+12h^2)=0

We get rid of parentheses

-12h^2+96h-140=0

a = -12; b = 96; c = -140;
Δ = b2-4ac
Δ = 962-4·(-12)·(-140)
Δ = 2496
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:

$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-8\sqrt{39}}{2*-12}=\frac{-96-8\sqrt{39}}{-24}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+8\sqrt{39}}{2*-12}=\frac{-96+8\sqrt{39}}{-24}
$