I solved Isosceles Triangle if we know an Median on Leg and a Perimeter. But the solution seems excessively complicated to be computed other than by a computer.
I start with the following cubic equation. There are two valid results, both of which are for leg:
$4Px^3 − (5P^2 + m^2)x^2 + 2P^3x − (1/4)P^4 = 0$
where P is Perimeter and m is median on leg.
Is it possible either to find x in a easier way, or to calculate the problem in a completely different way?
I don't see from where comes your equation.
Anyway: there are infinitely many triangle that satisfy the given condition, depending on the length of the side $AB$ where you put the midpoint.
I suggest an analytic approach.
Let $A=(-a,0)$ and $B=(a,0)$ so that the mid point is $M=(0,0)$. If $P=(x,y)$ is the other vertex of the triangle, we want: $PM=m$ so : $P=(x,\sqrt{m^2-x^2})$ ( here I ignore the symmetric solution) .
So we have: $$ PA=\sqrt{(x+a)^2+(m^2-x^2)}=\sqrt{a^2+m^2+2ax} $$ and, analogously: $$ PB=\sqrt{a^2+m^2-2ax} $$ and, if $p$ is the perimeter, we find the equation: $$ p=PA+PB+AB \iff \sqrt{a^2+m^2+2ax}+\sqrt{a^2+m^2-2ax}+2a=p $$
that is not so difficult to solve.
squaring the equation $$ \sqrt{a^2+m^2+2ax}+\sqrt{a^2+m^2-2ax}=p-2a $$ we have: $$ a^2+m^2+2ax+a^2+m^2-2ax+2\sqrt{(a^2+m^2+2ax)(a^2+m^2-2ax)}=(p-2a)^2 $$ $$ 2(a^2+m^2)+2\sqrt{(a^2+m^2)^2-4a^2x^2}=(p-2a)^2 $$ $$ 2\sqrt{(a^2+m^2)^2-4a^2x^2}=(p-2a)^2-2(a^2+m^2) $$ squaring again you find a second degree (pure) equation in $x$: $$ 4\left[(a^2+m^2)^2-4a^2x^2\right]=\left[(p-2a)^2-2(a^2+m^2)\right]^2 $$ can you solve this, and discuss the solutions?