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7p^2+20p-3=0
a = 7; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·7·(-3)
Δ = 484
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{484}=22$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-22}{2*7}=\frac{-42}{14} =-3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+22}{2*7}=\frac{2}{14} =1/7 $
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