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3p^2+7p-7=0
a = 3; b = 7; c = -7;
Δ = b2-4ac
Δ = 72-4·3·(-7)
Δ = 133
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{133}}{2*3}=\frac{-7-\sqrt{133}}{6} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{133}}{2*3}=\frac{-7+\sqrt{133}}{6} $
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