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36h^2+13h+1=0
a = 36; b = 13; c = +1;
Δ = b2-4ac
Δ = 132-4·36·1
Δ = 25
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{25}=5$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*36}=\frac{-18}{72} =-1/4 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*36}=\frac{-8}{72} =-1/9 $
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