## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$$x=-5+\sqrt{37},\:x=-5-\sqrt{37}$$
To find the x-intercepts, we set the function equal to zero. Doing this, we find: $$0\cdot \:2\left(x-2\right)=\frac{4x-2}{x-2}\cdot \:2\left(x-2\right)+\frac{x+4}{2}\cdot \:2\left(x-2\right) \\ 0=2\left(4x-2\right)+\left(x+4\right)\left(x-2\right)\\ x^2+10x-12=0\\ x_{1,\:2}=\frac{-10\pm \sqrt{10^2-4\cdot \:1\left(-12\right)}}{2\cdot \:1}\\ x=-5+\sqrt{37},\:x=-5-\sqrt{37}$$