3x(2+x)/3=X+4

Solution for 3x(2+x)/3=X+4 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

3x(x+2)-x*3-12=0

We multiply parentheses

3x^2+6x-x*3-12=0

Wy multiply elements

3x^2+6x-3x-12=0

We add all the numbers together, and all the variables

3x^2+3x-12=0

a = 3; b = 3; c = -12;
Δ = b2-4ac
Δ = 32-4·3·(-12)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{17}}{2*3}=\frac{-3-3\sqrt{17}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{17}}{2*3}=\frac{-3+3\sqrt{17}}{6}
$