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(P)=11.5P-0.1P^2-150
We move all terms to the left:
(P)-(11.5P-0.1P^2-150)=0
We get rid of parentheses
0.1P^2-11.5P+P+150=0
We add all the numbers together, and all the variables
0.1P^2-10.5P+150=0
a = 0.1; b = -10.5; c = +150;
Δ = b2-4ac
Δ = -10.52-4·0.1·150
Δ = 50.25
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{-(-10.5)-\sqrt{50.25}}{2*0.1}=\frac{10.5-\sqrt{50.25}}{0.2} $$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10.5)+\sqrt{50.25}}{2*0.1}=\frac{10.5+\sqrt{50.25}}{0.2} $
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