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7p^2+17p=0
a = 7; b = 17; c = 0;
Δ = b2-4ac
Δ = 172-4·7·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{289}=17$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-17}{2*7}=\frac{-34}{14} =-2+3/7 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+17}{2*7}=\frac{0}{14} =0 $
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