2h+h^2-98=0

Solution for 2h+h^2-98=0 equation:

2h+h^2-98=0

a = 1; b = 2; c = -98;
Δ = b2-4ac
Δ = 22-4·1·(-98)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{11}}{2*1}=\frac{-2-6\sqrt{11}}{2}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{11}}{2*1}=\frac{-2+6\sqrt{11}}{2}
$