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20x^2-35x=0
a = 20; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·20·0
Δ = 1225
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{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*20}=\frac{0}{40} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*20}=\frac{70}{40} =1+3/4 $
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