P=(10+x)(20-2x)

Solution for P=(10+x)(20-2x) equation:

=(10+P)(20-2P)

We move all terms to the left:

-((10+P)(20-2P))=0

We add all the numbers together, and all the variables

-((P+10)(-2P+20))=0

We multiply parentheses ..

-((-2P^2+20P-20P+200))=0


We calculate terms in parentheses: -((-2P^2+20P-20P+200)), so:

(-2P^2+20P-20P+200)

We get rid of parentheses

-2P^2+20P-20P+200

We add all the numbers together, and all the variables

-2P^2+200

Back to the equation:

-(-2P^2+200)


We get rid of parentheses

2P^2-200=0

a = 2; b = 0; c = -200;
Δ = b2-4ac
Δ = 02-4·2·(-200)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$P_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*2}=\frac{-40}{4}
=-10
$
$P_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*2}=\frac{40}{4}
=10
$