Реште пожалуйста уравнение

$\frac{ x^{2} -8x+12}{x-2} \geq 0 \\ x^{2} -8x+12=0 \\ D=64-48=16 \\ \sqrt{D} =4 \\ x_{1} = \frac{8+4}{2} = \frac{12}{2} =6 \\ x_{2} = \frac{8-4}{2} = \frac{4}{2} =2 \\ x-2 \neq 0 \\ x \neq 2 \\ \\ \frac{(x-6)(x-2)}{(x-2)} \geq 0 \\$
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---------------------|--------------------------------|----------------> x
2 6

$x\in [6;+ \infty})$

$\sqrt{2x+1} =7-x \\ 2x+1=(7-x) ^{2} \\ 7-x \geq 0 \\ -x \geq -7 \\ x \leq 7 \\ \\ 2x+1=49+ x^{2} -14x \\ 2x+14x- x^{2} -49+1=0 \\ - x^{2} +16x-48=0 \\ x^{2} -16x+48=0 \\ D=256-192=64 \ \ \ \ \sqrt{D} =8 \\ x_{1} = \frac{16+8}{2} = \frac{24}{2} =12 \\ x_{2} = \frac{16-8}{2} =4 \\ Otviet:4$