Find all real solutions of the equation

**Solution**

Introduce y = x^2+2ax+\frac{1}{16}. Then squaring the right-hand side gives x = y^2+2ay+\frac{1}{16}. But

and

are inverse function so that their graphs are reflections of each other in the main diagonal y = x. Solutions to the original equation lie on both graphs and, hence, on the diagonal. It follows that any such solution also satisfies

This is a quadratic equation

with two solutions

For 0 < a < \frac{1}{4}, -1 < (2a-1) < -\frac{1}{2}, so that

which means that the expression under the radical is positive and the roots are real.

To make the verification easier set b = \frac{1-2a}{2} and observe that then a = \frac{1-2b}{2}. In this notations,