\(\dfrac{x^2-3x+2}{x^2+4x+5}=\dfrac{1}{2}\Rightarrow2\left(x^2-3x+2\right)=x^2+4x+5\)

\(\Leftrightarrow2x^2-6x+4=x^2+4x+5\)

\(\Leftrightarrow x^2-10x-1=0\)

=> We have two solutions : \(\left[{}\begin{matrix}x=5+\sqrt{26}\\x=5-\sqrt{26}\end{matrix}\right.\)

We have : x2−3x+2x2+4x+5=12

⇒(x2−3x+2).2=x2+4x+5

⇔2x2−6x+4=x2+4x+5

⇔2x2−x2−6x−4x=5−4

⇔x2−10x−1=0

⇔x2−10x+25−26=0

​ ​⇔(x−5)2=26

=> x - 5 = ​ ​​√26

=> x = √26+5

Searching4You was right

We have : \(\dfrac{x^2-3x+2}{x^2+4x+5}=\dfrac{1}{2}\)

\(\Rightarrow\left(x^2-3x+2\right).2=x^2+4x+5\)

\(\Leftrightarrow2x^2-6x+4=x^2+4x+5\)

\(\Leftrightarrow2x^2-x^2-6x-4x=5-4\)

\(\Leftrightarrow x^2-10x-1=0\)

\(\Leftrightarrow x^2-10x+25-26=0\)

​ ​\(\Leftrightarrow\left(x-5\right)^2=26\)

=> x - 5 = ​ ​​\(\sqrt{26}\)

=> x = \(\sqrt{26}+5\)