I want to show using the $\epsilon$-$\delta$ criterion that the function $f(x)=-12x^2+17x+14$ is continuous on $\mathbb{R}$.

I have done the following:

Let $\epsilon>0$ be arbitrary.

For all $x$ with $|x-x\_0|<\delta$ we have that \begin{align\*}|f(x)-f(x\_0)|&=|(-12x^2+17x+14)-(-12x\_0^2+17x\_0+14)|\\ & =|-12(x^2-x\_0^2)+17(x-x\_0)|\\ & <12|x^2-x\_0^2|+17|x-x\_0|\\ & =12|(x-x\_0)(x+x\_0)|+17|x-x\_0|\\ & =12|x-x\_0|\cdot |x+x\_0|+17|x-x\_0|\\ & =(12\cdot |x+x\_0|+17)\cdot |x-x\_0|\\ & <(12\cdot |x+x\_0|+17)\cdot \delta \\ & =(12\cdot |x-x\_0+x\_0+x\_0|+17)\cdot \delta\\ & <(12\cdot |(x-x\_0)+2x\_0|+17)\cdot \delta\\ & <(12\cdot |x-x\_0|+12\cdot |2x\_0|+17)\cdot \delta\\ & <(12\cdot \delta+12\cdot |2x\_0|+17)\cdot \delta\\ & =12\cdot \delta^2+(24\cdot |x\_0|+17)\cdot \delta\end{align\*} We are looking for a $\delta$ such that \begin{align\*}12\cdot \delta^2+(24\cdot |x\_0|+17)\cdot \delta<\epsilon&\Rightarrow 12\cdot \delta^2+(24\cdot |x\_0|+17)\cdot \delta-\epsilon<0 \\ & \Rightarrow \delta^2+\frac{24\cdot |x\_0|+17}{12}\cdot \delta-\frac{\epsilon}{12}<0\end{align\*} We get the negative sign between if $\delta$ is between the two roots of the equation $\delta^2+\frac{24\cdot |x\_0|+17}{12}\cdot \delta-\frac{\epsilon}{12}=0$.

The roots are $$\delta\_{1,2}=-24|x\_0|-17\pm \sqrt{576|x\_0|^2+816|x\_0|+48\epsilon+289}$$ We have to consider also that $\delta$ must be positive, therefore we are looking for a $\delta$ in the interval $\left (0, -24|x\_0|-17+\sqrt{576|x\_0|^2+816|x\_0|+48\epsilon+289}\right )$ for example $\delta=\frac{-24|x\_0|-17+ \sqrt{576|x\_0|^2+816|x\_0|+48\epsilon+289}}{2}$.

Therefore, we have:

Let $\epsilon>0$ there there is a $\delta$, for example $\delta=\frac{-24|x\_0|-17+ \sqrt{576|x\_0|^2+816|x\_0|+48\epsilon+289}}{2}$ that for all $x\in \mathbb{R}$ with $|x-x\_0|<\delta$ it holds that $|f(x)-f(x\_0)|<\epsilon$.

Is everything correct? Could I improve something?