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3h^2+57h-2=0
a = 3; b = 57; c = -2;
Δ = b2-4ac
Δ = 572-4·3·(-2)
Δ = 3273
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(57)-\sqrt{3273}}{2*3}=\frac{-57-\sqrt{3273}}{6} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(57)+\sqrt{3273}}{2*3}=\frac{-57+\sqrt{3273}}{6} $
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