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h^2=99
We move all terms to the left:
h^2-(99)=0
a = 1; b = 0; c = -99;
Δ = b2-4ac
Δ = 02-4·1·(-99)
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{-(0)-6\sqrt{11}}{2*1}=\frac{0-6\sqrt{11}}{2} =-\frac{6\sqrt{11}}{2} =-3\sqrt{11} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{11}}{2*1}=\frac{0+6\sqrt{11}}{2} =\frac{6\sqrt{11}}{2} =3\sqrt{11} $
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