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3p^2+3p=0
a = 3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·3·0
Δ = 9
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{9}=3$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*3}=\frac{-6}{6} =-1 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*3}=\frac{0}{6} =0 $
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