$$\frac{\sqrt{4b-3}-1}{2}$$

$x = 1/2 (\sqrt{4 b + 1} + 1)$, $x = 1/2 (1 -\sqrt{4 b + 1})$, $x = 1/2 (\sqrt{4 b - 3} - 1)$, $x = 1/2 (-\sqrt{4 b - 3} - 1)$.

oralayhan wrote: $x = 1/2 (\sqrt{4 b + 1} + 1)$, $x = 1/2 (1 -\sqrt{4 b + 1})$, $x = 1/2 (\sqrt{4 b - 3} - 1)$, $x = 1/2 (-\sqrt{4 b - 3} - 1)$. extraneous solutions