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20x^2+9x-20=0
a = 20; b = 9; c = -20;
Δ = b2-4ac
Δ = 92-4·20·(-20)
Δ = 1681
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{1681}=41$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-41}{2*20}=\frac{-50}{40} =-1+1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+41}{2*20}=\frac{32}{40} =4/5 $
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