Evaluate: (x)/(x+4)=(2)/(x+2)-(2x)/(x^2+6x+8)

Expression: $x^{2}-3x=10$

Move $ 10 $to the left side

$x^{2}-3x-10=0$

Solve with the quadratic formula

$x_{1,2}=\frac{-(-3)\pm \sqrt{(-3)^{2}-4\cdot 1\cdot (-10)}}{2\cdot 1}$

$\sqrt{(-3)^{2}-4\cdot 1\cdot (-10)}=7$

$x_{1,2}=\frac{-(-3)\pm 7}{2\cdot 1}$

Separate the solutions

$x_{1}=\frac{-(-3)+7}{2\cdot 1},x_{2}=\frac{-(-3)-7}{2\cdot 1}$

$x=\frac{-(-3)+7}{2\cdot 1}:{\quad}5$

$=5$

$x=\frac{-(-3)-7}{2\cdot 1}:{\quad}-2$

$=-2$

The solutions to the quadratic equation are:

$x=5,x=-2$