2/3x-2(3+x)=-x-5

Solution for 2/3x-2(3+x)=-x-5 equation:

2/3x-2(3+x)=-x-5

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

2/3x-2(x+3)-(-1x-5)=0

We multiply parentheses

2/3x-2x-(-1x-5)-6=0

We get rid of parentheses

2/3x-2x+1x+5-6=0

We multiply all the terms by the denominator


-2x*3x+1x*3x+5*3x-6*3x+2=0

Wy multiply elements

-6x^2+3x^2+15x-18x+2=0

We add all the numbers together, and all the variables

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

a = -3; b = -3; c = +2;
Δ = b2-4ac
Δ = -32-4·(-3)·2
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{33}}{2*-3}=\frac{3-\sqrt{33}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*-3}=\frac{3+\sqrt{33}}{-6}
$