Help with algebra

$\frac{\left(x+4\right)}{\left(x+5\right)}=\frac{\left(x-4\right)}{\left(2x\right)}+\frac{\left(x+3\right)}{\left(x+5\right)}$

Multiply by LCM :

$2x\left(x+4\right)=\left(x-4\right)\left(x+5\right)+2x\left(x+3\right)$

$2x^2+8x=3x^2+7x-20$

Using quadratic formula to solve, 

$x_1=\frac{-\left(-1\right)+9}{2\cdot \:1},\:x_2=\frac{-\left(-1\right)-9}{2\cdot \:1}$

$x = 5, -4$

$\frac{\left(x+4\right)}{\left(x+5\right)}=\frac{\left(x-4\right)}{\left(2x\right)}+\frac{\left(x+3\right)}{\left(x+5\right)}, 2x\left(x+4\right)=\left(x-4\right)\left(x+5\right)+2x\left(x+3\right), x=5, x=-4$

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Thanks guys.....another slightly different approach

We can subtract the second term from the right side from both sides  and we get

(1) / (x + 5)  =(x -4) /(2x)

Cross-multiply

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

2x = x^2 + x -20

x^2  -x - 20   =   0

Factor

(x -5) (x + 4)   = 0

Set each factor to 0 and solve for x and we  get  that   x= 5  or x  = -4