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25x^2+35x=0
a = 25; b = 35; c = 0;
Δ = b2-4ac
Δ = 352-4·25·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-35}{2*25}=\frac{-70}{50} =-1+2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+35}{2*25}=\frac{0}{50} =0 $
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