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3h^2-51=0
a = 3; b = 0; c = -51;
Δ = b2-4ac
Δ = 02-4·3·(-51)
Δ = 612
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{17}}{2*3}=\frac{0-6\sqrt{17}}{6} =-\frac{6\sqrt{17}}{6} =-\sqrt{17} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{17}}{2*3}=\frac{0+6\sqrt{17}}{6} =\frac{6\sqrt{17}}{6} =\sqrt{17} $
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