2(3x+8)=2x(x+3)

Solution for 2(3x+8)=2x(x+3) equation:

2(3x+8)=2x(x+3)

We move all terms to the left:

2(3x+8)-(2x(x+3))=0

We multiply parentheses

6x-(2x(x+3))+16=0


We calculate terms in parentheses: -(2x(x+3)), so:

2x(x+3)

We multiply parentheses

2x^2+6x

Back to the equation:

-(2x^2+6x)


We get rid of parentheses

-2x^2+6x-6x+16=0

We add all the numbers together, and all the variables

-2x^2+16=0

a = -2; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-2)·16
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{2}}{2*-2}=\frac{0-8\sqrt{2}}{-4}
=-\frac{8\sqrt{2}}{-4}
=-\frac{2\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{2}}{2*-2}=\frac{0+8\sqrt{2}}{-4}
=\frac{8\sqrt{2}}{-4}
=\frac{2\sqrt{2}}{-1}
$