Given equation is $(x-a)(x-b)=c, c \neq 0$
$\Rightarrow x^{2}-(a+b) x+a b-c=0$
Let $\alpha, \beta$ be the roots of this equation. $\alpha+\beta=a+b, \alpha \beta=a b-c$
Consider $(x-\alpha)(x-\beta)+c=0$
$\Rightarrow x^{2}-(\alpha+\beta) x+\alpha \beta+c=0$
Roots of this equation is
$x=\frac{(\alpha+\beta) \pm \sqrt{(\alpha+\beta)^{2}-4(\alpha \beta+c)}}{2}$
$=\frac{(a+b) \pm \sqrt{(a+b)^{2}-4(a b-c+c)}}{2}$
$=\frac{(a+b) \pm \sqrt{a^{2}+b^{2}-2 a b}}{2}$
$=\frac{(a+b) \pm \sqrt{(a-b)^{2}}}{2}=\frac{(a+b) \pm(a-b)}{2}$
$=\frac{a+b+a-b}{2}, \frac{a+b-a+b}{2}=a, b$.
Hence, roots of the equation $(x-\alpha)(x-\beta)+\mathrm{c}=0$ are $a$ and $b$.