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(2p+3)(3p+1)=18
We move all terms to the left:
(2p+3)(3p+1)-(18)=0
We multiply parentheses ..
(+6p^2+2p+9p+3)-18=0
We get rid of parentheses
6p^2+2p+9p+3-18=0
We add all the numbers together, and all the variables
6p^2+11p-15=0
a = 6; b = 11; c = -15;
Δ = b2-4ac
Δ = 112-4·6·(-15)
Δ = 481
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{-(11)-\sqrt{481}}{2*6}=\frac{-11-\sqrt{481}}{12} $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{481}}{2*6}=\frac{-11+\sqrt{481}}{12} $
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