18=399+(5p)(20+p)

Solution for 18=399+(5p)(20+p) equation:

18=399+(5p)(20+p)

We move all terms to the left:

18-(399+(5p)(20+p))=0

We add all the numbers together, and all the variables

-(399+5p(p+20))+18=0


We calculate terms in parentheses: -(399+5p(p+20)), so:

399+5p(p+20)

determiningTheFunctionDomain
5p(p+20)+399

We multiply parentheses

5p^2+100p+399

Back to the equation:

-(5p^2+100p+399)


We get rid of parentheses

-5p^2-100p-399+18=0

We add all the numbers together, and all the variables

-5p^2-100p-381=0

a = -5; b = -100; c = -381;
Δ = b2-4ac
Δ = -1002-4·(-5)·(-381)
Δ = 2380
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:

$\sqrt{\Delta}=\sqrt{2380}=\sqrt{4*595}=\sqrt{4}*\sqrt{595}=2\sqrt{595}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-2\sqrt{595}}{2*-5}=\frac{100-2\sqrt{595}}{-10}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+2\sqrt{595}}{2*-5}=\frac{100+2\sqrt{595}}{-10}
$