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20q^2+19q+3=0
a = 20; b = 19; c = +3;
Δ = b2-4ac
Δ = 192-4·20·3
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*20}=\frac{-30}{40} =-3/4 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*20}=\frac{-8}{40} =-1/5 $
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