0=2h^2+12h-4986

Solution for 0=2h^2+12h-4986 equation:

0=2h^2+12h-4986

We move all terms to the left:

0-(2h^2+12h-4986)=0

We add all the numbers together, and all the variables

-(2h^2+12h-4986)=0

We get rid of parentheses

-2h^2-12h+4986=0

a = -2; b = -12; c = +4986;
Δ = b2-4ac
Δ = -122-4·(-2)·4986
Δ = 40032
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:

$\sqrt{\Delta}=\sqrt{40032}=\sqrt{144*278}=\sqrt{144}*\sqrt{278}=12\sqrt{278}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12\sqrt{278}}{2*-2}=\frac{12-12\sqrt{278}}{-4}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12\sqrt{278}}{2*-2}=\frac{12+12\sqrt{278}}{-4}
$