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

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

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

We move all terms to the left:

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


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We add all the numbers together, and all the variables

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

We multiply parentheses

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

-4x^2+2x^2+10x-12x+3=0

We add all the numbers together, and all the variables

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

a = -2; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-2)·3
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-2}=\frac{2-2\sqrt{7}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-2}=\frac{2+2\sqrt{7}}{-4}
$