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20x^2-29x+9=0
a = 20; b = -29; c = +9;
Δ = b2-4ac
Δ = -292-4·20·9
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-11}{2*20}=\frac{18}{40} =9/20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+11}{2*20}=\frac{40}{40} =1 $
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