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20h^2+13h+2=0
a = 20; b = 13; c = +2;
Δ = b2-4ac
Δ = 132-4·20·2
Δ = 9
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{9}=3$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*20}=\frac{-16}{40} =-2/5 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*20}=\frac{-10}{40} =-1/4 $
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