S=x(40-2x)+x(40-2x)+x(80-2x)+x(80-2x)+(80-2x)(40-2x)

Solution for S=x(40-2x)+x(40-2x)+x(80-2x)+x(80-2x)+(80-2x)(40-2x) equation:

=S(40-2S)+S(40-2S)+S(80-2S)+S(80-2S)+(80-2S)(40-2S)

We move all terms to the left:

-(S(40-2S)+S(40-2S)+S(80-2S)+S(80-2S)+(80-2S)(40-2S))=0

We add all the numbers together, and all the variables

-(S(-2S+40)+S(-2S+40)+S(-2S+80)+S(-2S+80)+(-2S+80)(-2S+40))=0

We multiply parentheses ..

-(S(-2S+40)+S(-2S+40)+S(-2S+80)+S(-2S+80)+(+4S^2-80S-160S+3200))=0


We calculate terms in parentheses: -(S(-2S+40)+S(-2S+40)+S(-2S+80)+S(-2S+80)+(+4S^2-80S-160S+3200)), so:

S(-2S+40)+S(-2S+40)+S(-2S+80)+S(-2S+80)+(+4S^2-80S-160S+3200)

determiningTheFunctionDomain
(+4S^2-80S-160S+3200)+S(-2S+40)+S(-2S+40)+S(-2S+80)+S(-2S+80)

We multiply parentheses

(+4S^2-80S-160S+3200)-2S^2-2S^2-2S^2-2S^2+40S+40S+80S+80S

We get rid of parentheses

4S^2-2S^2-2S^2-2S^2-2S^2-80S-160S+40S+40S+80S+80S+3200

We add all the numbers together, and all the variables

-4S^2+3200

Back to the equation:

-(-4S^2+3200)


We get rid of parentheses

4S^2-3200=0

a = 4; b = 0; c = -3200;
Δ = b2-4ac
Δ = 02-4·4·(-3200)
Δ = 51200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{51200}=\sqrt{25600*2}=\sqrt{25600}*\sqrt{2}=160\sqrt{2}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160\sqrt{2}}{2*4}=\frac{0-160\sqrt{2}}{8}
=-\frac{160\sqrt{2}}{8}
=-20\sqrt{2}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160\sqrt{2}}{2*4}=\frac{0+160\sqrt{2}}{8}
=\frac{160\sqrt{2}}{8}
=20\sqrt{2}
$