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15-12p=3p^2
We move all terms to the left:
15-12p-(3p^2)=0
determiningTheFunctionDomain -3p^2-12p+15=0
a = -3; b = -12; c = +15;
Δ = b2-4ac
Δ = -122-4·(-3)·15
Δ = 324
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{324}=18$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-18}{2*-3}=\frac{-6}{-6} =1 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+18}{2*-3}=\frac{30}{-6} =-5 $
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