4h^2+2h-10=0

Solution for 4h^2+2h-10=0 equation:

4h^2+2h-10=0

a = 4; b = 2; c = -10;
Δ = b2-4ac
Δ = 22-4·4·(-10)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{41}}{2*4}=\frac{-2-2\sqrt{41}}{8}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{41}}{2*4}=\frac{-2+2\sqrt{41}}{8}
$