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9p^2+6p+1=12
We move all terms to the left:
9p^2+6p+1-(12)=0
We add all the numbers together, and all the variables
9p^2+6p-11=0
a = 9; b = 6; c = -11;
Δ = b2-4ac
Δ = 62-4·9·(-11)
Δ = 432
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}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12\sqrt{3}}{2*9}=\frac{-6-12\sqrt{3}}{18} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12\sqrt{3}}{2*9}=\frac{-6+12\sqrt{3}}{18} $
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