2h^2+2h-8/7=0

Solution for 2h^2+2h-8/7=0 equation:

2h^2+2h-8/7=0

We multiply all the terms by the denominator


2h^2*7+2h*7-8=0

Wy multiply elements

14h^2+14h-8=0

a = 14; b = 14; c = -8;
Δ = b2-4ac
Δ = 142-4·14·(-8)
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{161}}{2*14}=\frac{-14-2\sqrt{161}}{28}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{161}}{2*14}=\frac{-14+2\sqrt{161}}{28}
$