6.157. $2(x-1)^{2}-5(x-1)(x-a)+2(x-a)^{2}=0$.
## Solution.

Dividing the equation by $(x-a)^{2} \neq 0$, we have

$$
2\left(\frac{x-1}{x-a}\right)^{2}-5\left(\frac{x-1}{x-a}\right)+2=0
$$

Let $\frac{x-1}{x-a}=y$.

The equation in terms of $y$ becomes $2 y^{2}-5 y+2=0$, from which $y_{1}=\frac{1}{2}, y_{2}=2$.

Then $\frac{x-1}{x-a}=\frac{1}{2}, x_{1}=2-a$; or $\frac{x-1}{x-a}=2, x_{2}=2 a-1$.

If $x-a=0$, then $a-1=0$ and $x=a=1$.

Answer: $x_{1}=2-a ; x_{2}=2 a-1$.