12h^2-96h+140=0

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

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}
$