(1/4-a )-(1/4-a^2)=-0.5

You've lost a sign in there Anonymous!

Write this as 

1/4 - a - 1/4 + a² = -1/2   remembering that two minuses make a plus, which is why the a² term is + not -.

Simplify:

a² - a = -1/2

Add 1/2 to both sides

a² - a + 1/2 = 0

Solve using the quadratic formula:

$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{\left({i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{a}} = {\frac{\left({i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\right)\\
{\mathtt{a}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\\
\end{array} \right\}$$

You can see that the solutions are complex (that is, they involve i, the square root of -1); there are no real- number solutions.

. 

$$\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,-\,}}{\mathtt{a}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,-\,}}{{\mathtt{a}}}^{{\mathtt{2}}}\right) = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$    |*2
$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}} = -{\mathtt{1}}$$$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{\left({i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{a}} = {\frac{\left({i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\right)\\
{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\\
\end{array} \right\}$$

just one xtra thing to solve this without you need

 standart quadratic formula enjoy