Let $\alpha ,\beta$   be the roots of the equation $\left(x-a\right)\left(x-b\right)=c,c\ne 0$ . Then the roots of the equation $\left(x-\alpha \right)\left(x-\beta \right)+c=0$  are

$x^2-bx-ax+ab-c=0$

$x^2-\left(a+b\right)x+ab-c=0$

$\alpha +\beta =a+b,\alpha \beta =ab-c$

$\left(x-\alpha \right)\left(x-\beta \right)+c=0$

$x^2-\beta x-\alpha x+\alpha \beta +c=0$

$x^2-\left(\alpha +\beta \right)x+\alpha \beta +c=0$

$x^2-\left(a+b\right)x+ab-c+c=0$

$x^2-\left(a+b\right)x+ab=0$

$\therefore$  roots are $a,b$