h^2+28h-4=0

Solution for h^2+28h-4=0 equation:

h^2+28h-4=0

a = 1; b = 28; c = -4;
Δ = b2-4ac
Δ = 282-4·1·(-4)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-20\sqrt{2}}{2*1}=\frac{-28-20\sqrt{2}}{2}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+20\sqrt{2}}{2*1}=\frac{-28+20\sqrt{2}}{2}
$