Solve $x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}} $

Hint

$$x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\\ (x+a)^2-a^2+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\\ (x+a)^2=a^2-\frac{1}{16}-a+\sqrt{a^2+x-\frac{1}{16}}\\ (x+a)^2+(x+a)=a^2+x-\frac{1}{16}+\sqrt{a^2+x-\frac{1}{16}}$$

Now call:

$$y=\sqrt{a^2+x-\frac{1}{16}}\to y^2=a^2+x-\frac{1}{16}$$

and then

$$(x+a)^2+(x+a)=y^2+y\\ (x+a)^2-y^2+(x+a)-y=0\\ (x+a-y)(x+a+y+1)=0$$

Can you finish?

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

$$x=\displaystyle\frac{1}{2}-a \pm \sqrt{\displaystyle\frac{3}{16}- a (1-a)}$$