Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \sqrt{m^2 + 4c}}{2}$, $B_x = \frac{m - \sqrt{m^2 + 4c}}{2} $
$A_y = \frac{m^2 + m\sqrt{m^2 + 4c}}{2} + c$, $B_y = \frac{m^2 - m\sqrt{m^2 + 4c}}{2} + c$
using distance formula(not showing all steps)
$AP = \frac{m + \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$BP = \frac{m - \sqrt{m^2 + 4c}}{2}(\sqrt{m^2 + 1}) $
$AP - BP = 1$
$(\sqrt{m^2 + 4c})(\sqrt{m^2 + 1}) = 1$
$m^4 + m^2(4c + 1) + 4c - 1 = 0$
well I could manipulate this into quadratic but that doesn't really help me with coefficient with c.

Since you assume that $m > 0,$ this result of your calculations is good:
$$ AP = \frac{m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}. \tag1$$
Here's where you get in a bit of trouble:
$$ BP \stackrel?= \frac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}. \tag2$$
You want $AP - BP = 1,$ and I think the best interpretation of the problem statement  interprets $AP - BP$ as the difference of two positive lengths (rather than a negative length subtracted from a positive length). Moreover, $AP$ must be the greater of the two lengths in order for the difference to be positive.
The problem with Equation $(2)$ is that if $c > 0$ then the expression on the right side of the equation is negative. A better equation is:
$$ BP = \left\lvert \frac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}\right\rvert.$$
A more useful correct equation is
$$ BP = \begin{cases}
 \dfrac{-m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1} & c \geq 0, \\[1ex]
 \dfrac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1} & c < 0.
\end{cases} \tag3$$
The $c < 0$ case still looks shaky because of the (apparent) possibility that
$m^2 + 4c < 0,$ which would make the square root undefined, but what actually happens is that for very large negative $c$ the value of $m$ also will be large.
Equation $(1)$, on the other hand, is good because with $m > 0$ you are guaranteed that the expression on the right-hand side of the equation is positive,
and because the expression on the right-hand side is larger than either of the two expressions on the right-hand side of Equation $(3)$,
so you have chosen the correct expression for $AP$ in either case.
The two cases in Equation $(3)$ can (and I think should) be considered separately.
In the $c \geq 0$ case we have
\begin{align}
1 &= AP - BP \\
&= \frac{m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}
 - \frac{-m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1} \\
&= m \sqrt{m^2 + 1} 
\end{align}
and therefore
$$ m^4 + m^2 - 1 = 0, $$
for which the only solution (since $m^2$ must be positive) is
$$ m^2 = \frac12(\sqrt5 - 1). $$
In the $c < 0$ case, on the other hand, your further calculations are correct, and $m^2$ is the positive root $v$ of the quadratic equation
$$ v^2 + (4c + 1) v + 4c - 1 = 0, $$
that is,
\begin{align}
 m^2 &= \frac{-(4c + 1) + \sqrt{(4c + 1)^2 - 4(4c - 1)}}{2} \\
     &= \frac{-4c - 1 + \sqrt{(4c - 1)^2 + 4}}{2}.
\end{align}
You cannot eliminate $c$ from the solution in this case because the slope of the line actually does depend on how negative $c$ is.
With a $y$-intercept very far down the negative $y$ axis you need a steep slope in order to intersect the parabola.
My hunch is that you were supposed to solve the case $c \geq 0.$
This could have been stated explicitly, or it could have been implied by stating that $P$ is between $A$ and $B.$
