How can I find the eigenvalues of a matrix without a factorable trinomial?

I need to find the eigenvalues/eigenvectors of the following matrix:

$$A= \begin{bmatrix}-26 & 20\\-30 & 24\end{bmatrix}$$

I first used the equation for eigenvectors, and sorted out the determinant:

$$P_A(x) = \det (xI - A)$$ $$= \det\begin{bmatrix}x & 0\\0 & x\end{bmatrix} - \begin{bmatrix}-26 & 20\\-30 & 24\end{bmatrix}$$ $$= \det\begin{bmatrix}x + 20 & -20\\ 30 & x - 24\end{bmatrix}$$ $$= (x + 20)(x - 24) - (-600)$$ $$= x^2 - 4x + 120$$

The problem at this point, is that the trinomial $x^2 - 4x + 120$ can't be factored into $(x + a)(x + b)$. At least I don't think. The computer on Wolfram hasn't been able to do it, so I don't think it's factorable, which would mean I can't find the two x values that would give me the eigenvalues.

So how can I get the x values from a trinomial that isn't factoable? How else could I find the matrix's eigenvalues?

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You're asking how to find the roots of a quadratic polynomial, right? By the way, one of your 26's mysteriously turned into a 20. – Hurkyl Nov 13 '12 at 23:22
It should be $$\det\begin{bmatrix}x + 26 & -20\\ 30 & x - 24\end{bmatrix}$$ – Américo Tavares Nov 13 '12 at 23:29

You didn't compute the polynomial correctly, it should be $x^2+2x-24 = (x+6)(x-4)$.
Completing the square, we see that $$x^2-4x+120=x^2-4x+4+116=(x-2)^2+116,$$ so $x^2-4x+4=0$ if and only if $(x-2)^2=-116$ if and only if $x-2=\pm i\sqrt{116}=\pm 2i\sqrt{29}$ if and only if $x=2\pm 2i\sqrt{29}.$ Thus, we can factor it as $$\left(x-2-2i\sqrt{29}\right)\left(x-2+2i\sqrt{29}\right).$$ However, you've miscalculated the characteristic polynomial. It is indeed factorable.