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35n+49n^2=0
a = 49; b = 35; c = 0;
Δ = b2-4ac
Δ = 352-4·49·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1225}=35$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-35}{2*49}=\frac{-70}{98} =-5/7 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+35}{2*49}=\frac{0}{98} =0 $
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