3x+4x+5x=(2+x)*(2+x)

Solution for 3x+4x+5x=(2+x)*(2+x) equation:

3x+4x+5x=(2+x)(2+x)

We move all terms to the left:

3x+4x+5x-((2+x)(2+x))=0

We add all the numbers together, and all the variables

3x+4x+5x-((x+2)(x+2))=0

We add all the numbers together, and all the variables

12x-((x+2)(x+2))=0

We multiply parentheses ..

-((+x^2+2x+2x+4))+12x=0


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

(+x^2+2x+2x+4)

We get rid of parentheses

x^2+2x+2x+4

We add all the numbers together, and all the variables

x^2+4x+4

Back to the equation:

-(x^2+4x+4)


We add all the numbers together, and all the variables

12x-(x^2+4x+4)=0

We get rid of parentheses

-x^2+12x-4x-4=0

We add all the numbers together, and all the variables

-1x^2+8x-4=0

a = -1; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·(-1)·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{3}}{2*-1}=\frac{-8-4\sqrt{3}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{3}}{2*-1}=\frac{-8+4\sqrt{3}}{-2}
$