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7r^2-35r=0
a = 7; b = -35; c = 0;
Δ = b2-4ac
Δ = -352-4·7·0
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1225}=35$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-35}{2*7}=\frac{0}{14} =0 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+35}{2*7}=\frac{70}{14} =5 $
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