P(x)=(125+5x)(100-2x)-(500+30(100-2x))

Solution for P(x)=(125+5x)(100-2x)-(500+30(100-2x)) equation:

(P)=(125+5P)(100-2P)-(500+30(100-2P))

We move all terms to the left:

(P)-((125+5P)(100-2P)-(500+30(100-2P)))=0

We add all the numbers together, and all the variables

P-((5P+125)(-2P+100)-(500+30(-2P+100)))=0

We multiply parentheses ..

-((-10P^2+500P-250P+12500)-(500+30(-2P+100)))+P=0


We calculate terms in parentheses: -((-10P^2+500P-250P+12500)-(500+30(-2P+100))), so:

(-10P^2+500P-250P+12500)-(500+30(-2P+100))

We get rid of parentheses

-10P^2+500P-250P-(500+30(-2P+100))+12500


We calculate terms in parentheses: -(500+30(-2P+100)), so:

500+30(-2P+100)

determiningTheFunctionDomain
30(-2P+100)+500

We multiply parentheses

-60P+3000+500

We add all the numbers together, and all the variables

-60P+3500

Back to the equation:

-(-60P+3500)


We add all the numbers together, and all the variables

-10P^2+250P-(-60P+3500)+12500

We get rid of parentheses

-10P^2+250P+60P-3500+12500

We add all the numbers together, and all the variables

-10P^2+310P+9000

Back to the equation:

-(-10P^2+310P+9000)


We get rid of parentheses

10P^2-310P+P-9000=0

We add all the numbers together, and all the variables

10P^2-309P-9000=0

a = 10; b = -309; c = -9000;
Δ = b2-4ac
Δ = -3092-4·10·(-9000)
Δ = 455481
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:

$\sqrt{\Delta}=\sqrt{455481}=\sqrt{9*50609}=\sqrt{9}*\sqrt{50609}=3\sqrt{50609}$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-309)-3\sqrt{50609}}{2*10}=\frac{309-3\sqrt{50609}}{20}
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-309)+3\sqrt{50609}}{2*10}=\frac{309+3\sqrt{50609}}{20}
$