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20z^2-31z+12=0
a = 20; b = -31; c = +12;
Δ = b2-4ac
Δ = -312-4·20·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-1}{2*20}=\frac{30}{40} =3/4 $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+1}{2*20}=\frac{32}{40} =4/5 $
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