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27p^2-12=0
a = 27; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·27·(-12)
Δ = 1296
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{1296}=36$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*27}=\frac{-36}{54} =-2/3 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*27}=\frac{36}{54} =2/3 $
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