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If $f(x)=dfrac{2x+1}{x^2+ax+1}$. Find the value(s) of $a$ so that $f(x)$ is continuous on all real numbers $mathbb{R}$.

Here, I'll write my attempt. The denominator $x^2+ax+1 neq 0$. Then by using the quadratic formula, we obtain the following equation $dfrac{-a pm sqrt{a^2-4}}{2} neq 0$. That means $-a pm sqrt{a^2-4} neq 0$. Then $-a = pm sqrt{a^2-4}$. By squaring both sides, we obtain $a^2 neq a^2-4$ which is always true. So the conclusion is that $a$ is any real number. Your comment is highly appreciated.

The case you want is that $x^2+ax+1neq0$ on the real line, which is equivalent to saying that $x^2+ax+1$ has complex roots

you need to be very careful with your inequalities, since for example setting setting $a=-2$ yields that $x=2$ is a solution of your denominator. In particular you need not look for $neq 0$ but you need to look for $neq x$, since the existence of a soltion to your quadratic equation would yield to a zero of the denominator. Hence if you pick $a$ such that $a^2<4$ you are done, since then your squareroot does not exist in $mathbb{R}$ (pictorially your parabola does not touch the x axis), and so the fraction is defined everywhere.

You're interpreting the quadratic formula in a wrong way: asking for
$$
frac{-apmsqrt{a^2-4}}{2}ne0
$$
is imposing that the polynomial doesn't have $0$ among its roots, which is surely true for every $a$, because the product of the roots is $1$ (Viète's formulas). So no wonder you found that your inequation is satisfied for every value of $a$.

You rather want to impose that the polynomial $x^2+ax+1$ has no real root. Whenever $a^2ge4$ there are real roots given by the quadratic formula. So you want $a^-4<0$, that is $-2<a<2$.

A very simple counterexample would be choosing $a = 2$.

$$x^2+2x+1 neq 0 implies (x+1)^2 neq 0 implies xneq -1$$

Thus, $a in mathbb{R}$ isn’t correct.

Your error is that you used the quadratic formula incorrectly.

$$ax^2+bx+c = 0 implies color{blue}{x} = frac{-bpmsqrt{b^2-4ac}}{2a}$$

You used $color{red}{0}$ instead of $color{blue}{x}$ in your inequality.

$$x^2+ax+1 neq 0$$
$$x neq frac{-apmsqrt{a^2-4}}{2}$$

If you think about the function $f(x) = x^2+ax+1$, what value of $a$ would produce no solutions (a.k.a $x$-intercepts)? The discriminant must become negative.

$$Delta < 0 implies a^2-4 < 0 implies a^2 < 4 implies a < 2 text{ and } a > -2 implies a in (-2, 2)$$