values of $p$ that ensure quadratic formula has real solutions for $x$ So using the quadratic formula to give the solution for $x$ (in terms of $p$) for:
$$px^2+2x+1=0$$
which as far as i've figured out gives:
$$x = \frac{-2\pm\sqrt{4-4p}}{2p}$$
but i don't know how to find the values of $p$ that make this equation have real solutions for $x$.
 A: It has a real solution if the term inside the square root is nonnegative, that is $$4-4p \geq 0$$
$$4\geq 4p$$
Divided by $4$, we have
$$1 \geq p$$
Note that $4-4p$ the term inside the square root, is also known as the discriminant.
Remark:
Consider the case when $p=0$ separately, in that case, the equation is a linear equation.
A: The $Δ$(the discriminant of a quadratic)  of the quadratic must be non negative for the quadratic to have real solutions. 
In your case $$Δ=4-4p$$
Hence $$4-4p\ge 0$$
$$\Rightarrow p-1\le 0$$
$$\Rightarrow p\le 1$$
A: We have, $$x = \frac{-2\pm\sqrt{4-4p}}{2p} = \frac{-1\pm \sqrt{1-p}}{p}.$$ Since $p\in\mathbb{R}$ then $\sqrt{-1}\nmid p$ and therefore $p\not>1\Leftrightarrow p\leqslant 1$ since if on the other hand, $p > 0$, then $1 - p < 0$ and therefore $\sqrt{-1}\mid \sqrt{1 - p}$. In addition to that, $-1 = \sqrt{-1}^2$ thus the entire numerator is a multiple of $\sqrt{-1}$. This means that $\sqrt{-1}\mid px$, proving the claim that $p\leqslant 1$. Of course though, we also cannot divide by $0$, therefore $p< 0$.
This is why that for any $x\in\mathbb{R}$, if $ax^2 + bx + c = 0$ then the expression under the root in the quadratic formula, $$x = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a},$$ namely $(b^2 - 4ac)$, must be non-negative since $-b = b\sqrt{-1}^2$, therefore $b^2 - 4ac\geqslant 0\Leftrightarrow 4ac \geqslant b^2$.
In clearing up a potential misunderstanding:
I say that $p < 0$, but if $p=0$ then $\exists x \in \left\{\mathbb{R} : px^2 + 2x + 1 = 0\right\}\Leftrightarrow x = -\dfrac 12$. Although this is true, this implies that, $$px^2 + 2x + 1 = 0\Leftrightarrow 2x + 1 = 0,$$ which is why $x = -\dfrac 12$ however if $p = 0$ then the above equation is not a quadratic trinomial to begin with.
