P(x)=60x-0.4x^2-(2x+15)

Solution for P(x)=60x-0.4x^2-(2x+15) equation:

(P)=60P-0.4P^2-(2P+15)

We move all terms to the left:

(P)-(60P-0.4P^2-(2P+15))=0


We calculate terms in parentheses: -(60P-0.4P^2-(2P+15)), so:

60P-0.4P^2-(2P+15)

determiningTheFunctionDomain
-0.4P^2+60P-(2P+15)

We get rid of parentheses

-0.4P^2+60P-2P-15

We add all the numbers together, and all the variables

-0.4P^2+58P-15

Back to the equation:

-(-0.4P^2+58P-15)


We get rid of parentheses

0.4P^2-58P+P+15=0

We add all the numbers together, and all the variables

0.4P^2-57P+15=0

a = 0.4; b = -57; c = +15;
Δ = b2-4ac
Δ = -572-4·0.4·15
Δ = 3225
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:

$\sqrt{\Delta}=\sqrt{3225}=\sqrt{25*129}=\sqrt{25}*\sqrt{129}=5\sqrt{129}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-57)-5\sqrt{129}}{2*0.4}=\frac{57-5\sqrt{129}}{0.8}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-57)+5\sqrt{129}}{2*0.4}=\frac{57+5\sqrt{129}}{0.8}
$