Given $ax^2+bx+c=0$ with two real roots, $x_1>x_2$, find a quadratic equation whose roots are $x_1+1$ and $x_2-1$ without solving the first equation Roots of the equation $ (1): ax^2+bx+c=0$ are $x_{1}$ and $x_{2}$. They are both real.
Without solving first equation, make up new quadratic equation such that one of the its roots is $x_{1} + 1$ and second one is $x_{2}-1$. Note that $x_{2}>x_{1}$.
I started to solve this with Vieta's theorem but I couldn't continue. Deep explanation would be grateful!
Also I don't know what does not solve mean, I can't use a quadratic formula or something more specific?

$f(x)=ax^2+bx+c=a(x-x_1)(x-x_2)$ is given to you. Consider
\begin{align*}
g(x)&=a(x-(x_1+1))(x-(x_2-1))\\
&=a(x-x_1-1)(x-x_2+1)\\
&=a(\color{red}{(x-x_1)}-1)(\color{blue}{(x-x_2)}+1)\\
&=a(x-x_1)(x-x_2)+a(x-x_1)-a(x-x_2)-a\\
&=f(x)+a\underbrace{(x_2-x_1)}_{\text{given }>0}-a.
\end{align*}
Now $x_2-x_2=\sqrt{(x_1+x_2)^2-4x_1x_2}=\sqrt{\left(\frac{b^2}{a^2}\right)-\frac{4c}{a}}=\frac{\sqrt{b^2-4ac}}{a}$
Thus
$$g(x)=ax^2+bx+c+\sqrt{b^2-4ac}-a$$
