(16=2x)(8+2x)

Solution for (16=2x)(8+2x) equation:

(16=2x)(8+2x)

We move all terms to the left:

(16-(2x)(8+2x))=0

We add all the numbers together, and all the variables

(16-2x(2x+8))=0


We calculate terms in parentheses: +(16-2x(2x+8)), so:

16-2x(2x+8)

determiningTheFunctionDomain
-2x(2x+8)+16

We multiply parentheses

-4x^2-16x+16

Back to the equation:

+(-4x^2-16x+16)


We get rid of parentheses

-4x^2-16x+16=0

a = -4; b = -16; c = +16;
Δ = b2-4ac
Δ = -162-4·(-4)·16
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{2}}{2*-4}=\frac{16-16\sqrt{2}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{2}}{2*-4}=\frac{16+16\sqrt{2}}{-8}
$