Solve $x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $ 
Find the real roots of the equation $$x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $$ $$(0<a<\frac{1}{4})$$

My attempt,
$(x^2+2ax+\frac{1}{16}+a)^2=a^2+x-\frac{1}{16}$
I did an expansion which becomes  $x^4+4ax^3(4a^2+2a+\frac{1}{8})x^2+(4a^2+\frac{1}{4}a-1)x+\frac{1}{8}a+\frac{1}{16}+\frac{1}{256}=0$
$$\left [ x^4+(2a-1)x^3+\frac{1}{16}x^2 \right ]+\left [ (2a-1)x^3+(4a^2-1)x^2+(\frac{a}{8}+\frac{1}{16})x \right ] +\left [ (2a+\frac{17}{16})x^2+(4a^2+\frac{1}{8}a-\frac{17}{16})x+(\frac{1}{8}a+\frac{1}{16}+\frac{1}{256}) \right ]=0$$
$$\left [ x^2+(2a-1)x+\frac{1}{16} \right ]\left [ x^2(2a+1)x+(2a+\frac{17}{16}) \right ]=0$$
Since $0<a<\frac{1}{4}$, 
$x^2+(2a-1)x+\frac{1}{16}=0$
$x=\frac{1-2a}{2}\pm \sqrt{(\frac{1-2a}{2})^2-\frac{1}{16}}$
My solution is really messy and I'm unsure my answer.
My question: is my answer correct? And is there another way to solve this algebra question? Thanks in advance.

Note that the right-hand side is a solution to the equation
$$x=y^{2}+2 a y+\displaystyle\frac{1}{16}$$
Furthermore, the solution to this equation gives
$$y=x^{2}+2 a x+\displaystyle\frac{1}{16}$$
That is, we have an equation for points where a function equal to its inverse function. The points in such a case lie along the line $y=x$ (as can be verified in the plot), and we simply have a quadratic to solve:
$$x=x^{2}+2 a x+\displaystyle\frac{1}{16}$$
the solution to which is
$$x=\displaystyle\frac{1}{2}-a \pm \sqrt{\displaystyle\frac{3}{16}- a (1-a)}$$
