I will use a for $alpha$ and c for $cos(2*pi/5)$
If you use DeMoivre's and some trig, then you can show that
root1 = a + a^4 = 2cos(2*pi/5) = 2c
and root2 = a^2 + a^3 = 2cos(4*pi/5) = 2(2cos^2(2*pi/5) - 1) = 2(2c^2 - 1)
All the imaginary parts disappear.
So your quadratic is (x - root1)(x - root2) = (x - a - a^4)(x - a^2 - a^3)
Now, the constant parts of the quadratic eq must be equal, so
root1 * root2 = a^2 + a^4 + a^6 + a^7
We can show that a^2 + a^4 + a^6 + a^7 = -1
This means that
root1 * root2 = 2c * 2(2c^2-1) = -1
Now, solve for c. This will give you 3 solutions, but only one should make sense for cos(2*pi/5), and the problem already tells you the answer