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7p^2+7p-14=0
a = 7; b = 7; c = -14;
Δ = b2-4ac
Δ = 72-4·7·(-14)
Δ = 441
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{441}=21$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-21}{2*7}=\frac{-28}{14} =-2 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+21}{2*7}=\frac{14}{14} =1 $
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