# Solve for x if ( x-3) = 5/(2x+4)

### 3 Answers | Add Yours

We have to solve for x if (x-3) = 5/(2x+4).

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

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

=> 2x^2 - 6x + 4x - 12 = 5

=> 2x^2 - 2x - 17 = 0

Now the roots of this equation are:

x1 = [-b + sqrt(b^2 - 4ac)}/2a

=> [2 + sqrt(4 + 136)]/4

=> (2 + sqrt 140) / 4

=> 1/2 + sqrt 35/2

x2 = [-b - sqrt(b^2 - 4ac)}/2a

=> [2 - sqrt(4 + 136)]/4

=> (2 - sqrt 140) / 4

=> 1/2 - sqrt 35/2

**Therefore the values of x are 1/2 + sqrt 35/2 and 1/2 - sqrt 35/2.**

Given the equality ( x-3) = 5/(2x+4). We need to find x values that satisfies the equality.

==> (x-3) = 5/(2x+4).

First, we need to re-write the equation.

We will multiply by the denominator (2x+4).

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

Now we will open bracket.

==> x*2x + x*4 - 3*2x -3*4 = 5

==> 2x^2 + 4x - 6x - 12 = 5

Now we will subtract 5 from both sides.

==> 2x^2 -2x -12 - 5 = 0.

==> 2x^2 - 2x - 17 = 0

Now we will use the formula to determine the roots.

==> x1= (2 + sqrt(140) / 4 = 2+2sqrt35/ 4

==>** x1= (1/2)+ (sqrt35)/2**

**==> x2= (1/2)-(sqrt35)/2**

To solve for x if ( x-3) = 5/(2x+4).

We multiply both sides by (2x+4):

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

We expand the left and rewrite the equation:

2x^2+4x-6x-12= 5.

2x^2 - 2x-12-5 = 0.

2x^2 -2x -17 = 0.

Quadratic formula: the roots of ax^2+bx+c = 0 is given by:x1 = {-b+sqrt(b^2-4ac)}/2 and x2 = {-b-sqrt(b^2-4ac)}/2

We use the quadratic formula to get the roots:

x1 = {-(-2) +sqrt[(-2)^2-4*2(-17)]}/2*2 = (2+sqrt140}/4

x1 = {1+sqrt35}/2.

x2 = {1-sqrt35}/2.

Therefore the solutions of the equation are x1 = (1+sqrt35)/2 and x2 = (1-sqrt35)/2.