Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation 
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form 
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization?

For those amongst you folks, like egreg and myself, who enjoy completing the square, we can also have at it like this:
Set
$J = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}; \tag 1$
then our equation reads
$X^2 + X = J, \tag 2$
from which we may write, completing the square, 
$(X + \dfrac{1}{2}I)^2 = X^2 + X + \dfrac{1}{4} = J + \dfrac{1}{4}I = \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}; \tag 3$
our task is now to find the matrices $C$ such that
$C^2 = J + \dfrac{1}{4}I = \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}; \tag 4$
to this end we observe that, setting
$P = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \tag 5$
whence
$P^2 = I, \tag 6$
we have
$\left (\dfrac{1}{2}I + P \right )^2 = \dfrac{1}{4} I^2 + 2\left ( \dfrac{1}{2}I \right ) P + P^2 = \dfrac{1}{4} I + I + P =  \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}, \tag 7$
and also
$\left (I + \dfrac{1}{2} P \right )^2 = I^2 + 2I \left (\dfrac{1}{2}P \right ) + \dfrac{1}{4} P^2 = I + \dfrac{1}{4}I + P =  \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix}; \tag 8$
we are thus motivated to generalize by seeking solutions of the form $aI + bP$:
$(aI + bP)^2 = a^2 I^2 + 2abIP + b^2 P^2 = a^2 I + b^2 I + 2ab P$
$= (a^2 + b^2)I + 2ab P =  \begin{bmatrix} \dfrac{5}{4} & 1 \\ 1 & \dfrac{5}{4} \end{bmatrix} = \dfrac{5}{4}I + P; \tag 9$
thus,
$a^2 + b^2 = \dfrac{5}{4}; \; 2ab = 1; \tag{10}$
it follows from (10) that the ordered pairs $(a, b)$ lie on both the circle of radius $5/4$ centered at the origin in the $(a, b)$ plane, and also on the hypberbola $ab = 1/2$; since the points $\pm (1/\sqrt 2, 1/\sqrt 2)$ lie in the interior of the circle, the branches of this hyperbola cross the circle at exactly four points, and we have seen in (8)-(9) that two of them are $(1, 1/2)$ and $(1/2, 1)$; it is then evident from the symmetry $(a, b) \leftrightarrow (-a, -b)$ of the solutions to (10) that there are a total of precisely four pairs $(a, b)$ which solve (10) and hence (9), viz.
$(a, b) = \pm \left (1, \dfrac{1}{2} \right ), \; \pm \left ( \dfrac{1}{2}, 1 \right ); \tag{11}$
since
$X + \dfrac{1}{2}I = aI + bP = \pm \begin{bmatrix} 1 & \dfrac{1}{2} \\ \dfrac{1}{2} & 1 \end{bmatrix}, \pm \begin{bmatrix} \dfrac{1}{2} & 1 \\ 1 & \dfrac{1}{2} \end{bmatrix}, \tag{12}$
we have
$X = \begin{bmatrix}  \dfrac{1}{2} & \dfrac{1}{2} \\ \dfrac{1}{2} &  \dfrac{1}{2} \end{bmatrix}, \; \begin{bmatrix} -1 & -1 \\ -1 & -1  \end{bmatrix}, \; \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} -\dfrac{3}{2} & -\dfrac{1}{2} \\ -\dfrac{1}{2} & -\dfrac{3}{2} \end{bmatrix}, \tag{13}$
as the solutions to (2).
One question does, however, remain:  how do we know every $C$ as in (4) may be cast in the form $aI + bP$?  Well, as pointed out by Lord Shark the Unknown in his answer, $XJ = JX$ implies
$X = aI + bJ; \tag{14}$
but
$J = I + P, \tag{15}$
so in fact
$a I + b J = aI + b(I + P) = (a + b)I + bP, \tag{16}$
which shows that $X$ may be represented in the form $aI + bP$ as well; thus so may
$C = X + \dfrac{1}{2}I = (a + \dfrac{1}{2})I + bJ = (a + b + \dfrac{1}{2})I + bP. \tag{17}$
