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41h^2-14h=0
a = 41; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·41·0
Δ = 196
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{196}=14$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*41}=\frac{0}{82} =0 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*41}=\frac{28}{82} =14/41 $
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