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7h^2+3h-6=0
a = 7; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·7·(-6)
Δ = 177
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{-(3)-\sqrt{177}}{2*7}=\frac{-3-\sqrt{177}}{14} $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{177}}{2*7}=\frac{-3+\sqrt{177}}{14} $
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