Answer
No real solution
Work Step by Step
Use the quadratic formula after bringimg the equation to the standard form. $$
\begin{aligned}
\frac{2}{3} c^2-\frac{5}{6}&=\frac{1}{6} c-2\\
\left( \frac{2}{3} c^2-\frac{5}{6} \right)\cdot 6& =\left( \frac{1}{6} c-2\right) \cdot 6\\
4c^2-5 & =c-12 \\
4c^2-c-5+12 & = 0 \\
4c^2-c+7 & = 0 \\
\end{aligned}
$$ $$
\begin{aligned}
a & =4 \\
b& =-1\\
d & =7 \\
\end{aligned}
$$ $$
\begin{aligned}
c&= \frac{-b\pm \sqrt{b^2-4ad}}{2a}\\
c& =\frac{-(-1) \pm \sqrt{(-1)^2-4 \cdot 4 \cdot 7}}{2 \cdot 4}\\
& =\frac{1\pm\sqrt{-111}}{8}.
\end{aligned}
$$ The equation has no real solution because we cannot take the square root of a negative number.
