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25x-35x^2=0
a = -35; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·(-35)·0
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*-35}=\frac{-50}{-70} =5/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*-35}=\frac{0}{-70} =0 $
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