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14x^2+35x=0
a = 14; b = 35; c = 0;
Δ = b2-4ac
Δ = 352-4·14·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-35}{2*14}=\frac{-70}{28} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+35}{2*14}=\frac{0}{28} =0 $
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