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20x^2+39x+19=0
a = 20; b = 39; c = +19;
Δ = b2-4ac
Δ = 392-4·20·19
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-1}{2*20}=\frac{-40}{40} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+1}{2*20}=\frac{-38}{40} =-19/20 $
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