Quadratic equation with multiple parameter I have the following equation:
$$x^2 +mx+(a−m)^2 =0\text{ ($a$ be there as a fixed parameter)}$$
and I need to calculate for which values of $m$ the equation has exactly one solution for $x.$
I know I have to look the value of $m$ for which the discriminant of this equation is equal to $0.$ But every time when I went this path is got stuck.
Can someone give me some hint I need to search for or look for when trying to calculate this?

Addition:
One thing I forgot to mention is that first of all the goal is to find a value for $m$ and afterwards to calculate the solution(s) fo $x$.
 A: OK, calculate the discriminant:
$$m^2-4(a-m)^2=(m+2(a-m))(m-2(a-m))=0$$
This is zero precisely when one of the two factors is zero, so
$$m+2(a-m)=0\implies m=2a$$
$$m-2(a-m)=0\implies m=\frac23a$$
A: For $Ax^2+Bx+C=0$ to have only one solution, $B^2-4AC = 0$ or $B^2 = 4AC$.
So we need $m^2 = 4(a-m)^2 = (2(a-m))^2 \Rightarrow m = \pm(2a-2m) \Rightarrow \color{blue}{m =2a, \frac{2a}{3}}$

To calculate the solution of the equation,
$Ax^2+Bx+C$ has sum of the roots $= \frac{-B}{A}$. Here as the discriminant is zero, both the roots are equal, $x_1 = x_2$.
So, $2x_1 = -\frac{B}{A} = -m \Rightarrow \boxed{x_1 = x_2 =-\frac{m}{2}}$.
The roots are either $\boxed{-a}$ or $\boxed{-\frac{a}{3}}$ corresponding to $m=2a, \frac{2a}{3}$ respectively.

Graph for solutions:
Corresponding to $m=2a$ and $m=\frac{2a}{3}$
A: $$x^2+mx+(a-m)^2=0.$$
Since the roots are equal,
$$m^2-4(a-m)^2=0$$
$$m^2-(2a-2m)^2=0$$
$$(m+2a-2m)(m-2a+2m)=0$$
$$m=\frac{2a}{3},2a.$$
Plug in these values of m in original equation, we get two quadratic equations.
$$x^2+2ax+a^2=0$$
and
$$9x^2+6ax+a^2=0$$
Now you can find the roots of these equations by Vieta's formulas.
The roots of these equations are $-a$ and $\frac{-a}{3}$.
A: $A^2-B^2 =(A+B)(A-B)$, where $A=m$, $B=2(a-m)$. Did you get the hint?
A: As you said, the condition is that the discriminant is $0$.
In this case the discriminant is $-4 a^2 + 8 a m - 3 m^2$.
That's a quadratic in $m$: can you solve it?
A: If the equation $ \ x^2 + mx + (a−m)^2 \ = \ 0 \ \ $ has a single root, then the quadratic polynomial is a "binomial-square", which would have the form $ \ \left(x + \frac{m}{2} \right)^2 \ \ . $  Multiplying out the expressions in parentheses and putting the "constant terms" into correspondence, we obtain
$$ x^2 \ + \ mx \ + \ \frac{m^2}{4} \ \ = \ \ x^2 \ + \ mx \ + \ (a^2 - 2am + m^2) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ \frac{m^2}{4} \ \ = \ \ a^2 - 2am + m^2 \ \ \Rightarrow \ \ 3m^2 \ - \ 8am \ + \ 4a^2 \ \ = \ \ 0 \ \ . $$
This last equation factors as $ \ (3m - 2a)·(m - 2a) \ = \ 0 \ \ , $ so the possible values for $ \ m \ $ are $ \ m \ = \ \frac23·a \ \ , \ \ m \ = \ 2a \ \ . $
In answer to your appended question, writing the binomial-square makes it straightforward to see the solutions for $ \ x \ \ : $
$$ \left(x \ + \ \frac{\frac23·a}{2} \right)^2 \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ \ = \ \ -\frac13·a   $$
and
$$ \left(x \ + \ \frac{2·a}{2} \right)^2 \ \ = \ \ 0 \ \ \Rightarrow \ \ x \ \ = \ \ - a \ \ .  $$
