3h^2+26h-60=0

Solution for 3h^2+26h-60=0 equation:

3h^2+26h-60=0

a = 3; b = 26; c = -60;
Δ = b2-4ac
Δ = 262-4·3·(-60)
Δ = 1396
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:

$\sqrt{\Delta}=\sqrt{1396}=\sqrt{4*349}=\sqrt{4}*\sqrt{349}=2\sqrt{349}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{349}}{2*3}=\frac{-26-2\sqrt{349}}{6}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{349}}{2*3}=\frac{-26+2\sqrt{349}}{6}
$