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?
On
Since, $x_1 $ & $x_2$ are the roots of the quadratic equation: $ax^2+bx+c=0$, hence we have $$\text{sum of roots,} \ x_1+x_2=-\frac ba, \ \ \text{product of roots,}\ x_1x_2=\frac ca$$ Now, the equation having roots: $x_1+1 $ & $x_2-1$ is given as $$(x-(x_1+1))(x-(x_2-1))=0$$ $$x^2-(x_1+x_2)x+x_1x_2-1-(\underbrace{x_2-x_1}_{>0})=0$$ $$x^2-(x_1+x_2)x+x_1x_2-1-\sqrt{(x_2+x_1)^2-4x_1x_2}=0$$ setting the values, we get $$x^2-\left(-\frac ba\right)x+\frac ca-1-\sqrt{\left(-\frac ba\right)^2-4\frac ca}=0$$
$$\color{blue}{ax^2+bx+c-a+\sqrt{b^2-4ac}=0}$$ Above is the required quadratic equation.
$f(x)=ax^2+bx+c=a(x-x_1)(x-x_2)$ 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 $$g(x)=ax^2+bx+c+\sqrt{b^2-4ac}-a$$