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h^2=15
We move all terms to the left:
h^2-(15)=0
a = 1; b = 0; c = -15;
Δ = b2-4ac
Δ = 02-4·1·(-15)
Δ = 60
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{15}}{2*1}=\frac{0-2\sqrt{15}}{2} =-\frac{2\sqrt{15}}{2} =-\sqrt{15} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{15}}{2*1}=\frac{0+2\sqrt{15}}{2} =\frac{2\sqrt{15}}{2} =\sqrt{15} $
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