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3p^2+6p-9=0
a = 3; b = 6; c = -9;
Δ = b2-4ac
Δ = 62-4·3·(-9)
Δ = 144
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{144}=12$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12}{2*3}=\frac{-18}{6} =-3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12}{2*3}=\frac{6}{6} =1 $
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