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20x^2-65x+15=0
a = 20; b = -65; c = +15;
Δ = b2-4ac
Δ = -652-4·20·15
Δ = 3025
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{3025}=55$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-55}{2*20}=\frac{10}{40} =1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+55}{2*20}=\frac{120}{40} =3 $
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