320=140+(20-2x)(16-2x)

Solution for 320=140+(20-2x)(16-2x) equation:

320=140+(20-2x)(16-2x)

We move all terms to the left:

320-(140+(20-2x)(16-2x))=0

We add all the numbers together, and all the variables

-(140+(-2x+20)(-2x+16))+320=0

We multiply parentheses ..

-(140+(+4x^2-32x-40x+320))+320=0


We calculate terms in parentheses: -(140+(+4x^2-32x-40x+320)), so:

140+(+4x^2-32x-40x+320)

determiningTheFunctionDomain
(+4x^2-32x-40x+320)+140

We get rid of parentheses

4x^2-32x-40x+320+140

We add all the numbers together, and all the variables

4x^2-72x+460

Back to the equation:

-(4x^2-72x+460)


We get rid of parentheses

-4x^2+72x-460+320=0

We add all the numbers together, and all the variables

-4x^2+72x-140=0

a = -4; b = 72; c = -140;
Δ = b2-4ac
Δ = 722-4·(-4)·(-140)
Δ = 2944
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2944}=\sqrt{64*46}=\sqrt{64}*\sqrt{46}=8\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-8\sqrt{46}}{2*-4}=\frac{-72-8\sqrt{46}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+8\sqrt{46}}{2*-4}=\frac{-72+8\sqrt{46}}{-8}
$