6.123. Form a quadratic equation whose roots are one unit greater than the roots of the equation $a x^{2}+b x+c=0$.
Solution.

Let $y^{2}+p y+q=0$ be the desired equation with roots $y_{1}=x_{1}+1$, $y_{2}=x_{2}+1$. From the condition by Vieta's theorem

$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, \\
x_{1} \cdot x_{2}=\frac{c}{a},
\end{array}\right. \\
& \left\{\begin{array}{l}
y_{1}+y_{2}=x_{1}+1+x_{2}+1=x_{1}+x_{2}+2=-\frac{b}{a}+2=-p, \\
y_{1} \cdot y_{2}=\left(x_{1}+1\right)\left(x_{2}+1\right)=x_{1} x_{2}+x_{1}+x_{2}+1=\frac{c}{a}-\frac{b}{a}+1=q,
\end{array}\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
p=\frac{b-2 a}{a}, \\
q=\frac{a-b+c}{a}
\end{array}\right.
\end{aligned}
$$

We obtained the equation

$$
y^{2}+\frac{b-2 a}{a} y+\frac{a-b+c}{a}=0 \Leftrightarrow a y^{2}+(b-2 a) y+a-b+c=0
$$

Answer: $a y^{2}+(b-2 a) y+a-b+c=0$.