Solve x+1/6=1/2b^2x^2+bx+1/2

$$x + \frac { 1 } { 6 } = \frac { 1 } { 2 } b ^ { 2 } x ^ { 2 } + b x + \frac { 1 } { 2 }$$

Solve for b (complex solution)

$$b=\frac{\sqrt{3\left(6x+1\right)}-3}{3x}$$

$$b=-\frac{\sqrt{3}\left(\sqrt{6x+1}+\sqrt{3}\right)}{3x}\text{, }x\neq 0$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{\sqrt{3\left(b^{2}-6b+3\right)}+3b-3}{3b^{2}}\text{; }x=-\frac{-\sqrt{3\left(b^{2}-6b+3\right)}+3b-3}{3b^{2}}\text{, }&b\neq 0\\x=\frac{1}{3}\text{, }&b=0\end{matrix}\right.$$

Solve for b

$$b=-\frac{\sqrt{3}\left(\sqrt{6x+1}+\sqrt{3}\right)}{3x}$$

$$b=\frac{\sqrt{3\left(6x+1\right)}-3}{3x}\text{, }x\neq 0\text{ and }x\geq -\frac{1}{6}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{3\left(b^{2}-6b+3\right)}+3b-3}{3b^{2}}\text{; }x=-\frac{-\sqrt{3\left(b^{2}-6b+3\right)}+3b-3}{3b^{2}}\text{, }&\left(b\neq 0\text{ and }b\leq 3-\sqrt{6}\right)\text{ or }b\geq \sqrt{6}+3\\x=\frac{1}{3}\text{, }&b=0\end{matrix}\right.$$