I have the following differential equation
$$y''(t) + -2y'(t) + 5y(t) = 0$$
With the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
I can compute the solutions
$$a = ,1\quad b=-2,\quad c=5$$
$$\lambda_{1,2} = \frac{2 \pm \sqrt{4-4(1\cdot 5)}}{2} = 1 \pm \frac{\sqrt{-15}}{2} = 1 \pm i(\frac{\sqrt{15}}{2})$$
However I'm given the answer: $\lambda_{1,2} = 1 \pm i2$
What am I missing?
You're making a small arithmetic mistake.
$$\lambda_{1,2} = \frac{2 \pm \sqrt{4-4(1\cdot 5)}}{2} = 1 \pm \frac{\sqrt{-16}}{2} = 1 \pm i(\frac{4}{2}) = 1 \pm 2i$$
Because $4-4(1\cdot 5) = 4-20 = -16$