$f : [0, \infty ) \to \mathbb{R},$
that is, an expression for $f(x),$
such that for all $x,y \in [0,\infty)$
we have $f(x+2y)+2f^2(y)+x = f^2(x+2y)+2f(y)\frac{1}{4}.$

Determine the value of
$f$
for some simple cases. 
What about the value of
$f(0)?$

Why not try getting an equation of
$f$
with a single variable? 
… by substituting a suitable value for
$y.$

By setting
$x=y=0,$
$$\begin{aligned} f(0) + 2f^2(0) &= f^2(0) + 2f(0)  \frac{1}{4} \\ f^2(0)  f(0) + \frac{1}{4} &= 0 \\ \big(f(0)  \frac{1}{2}\big)^2 &= 0, \\ \end{aligned}$$
which gives us
$f(0) = \frac{1}{2}.$
Now, we can obtain$f(x)$
by setting$y=0:$
$$\begin{aligned} f(x)+2f^2(0)+x &= f^2(x)+2f(0)\frac{1}{4} \\ f^2(x)f(x)+\frac{1}{4}x &= 0.\\ \end{aligned}$$
Solving this quadratic equation gives us
$f(x) = \frac{1}{2} \pm \sqrt{x}.$