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14h^2+17h=0
a = 14; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·14·0
Δ = 289
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}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*14}=\frac{-34}{28} =-1+3/14 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*14}=\frac{0}{28} =0 $
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