Show that the roots of this quadratic equation $\displaystyle (k + 5)x^2 + 13x -6k = 0 $ are real and distinct for all the real values of k.

This was how I did it.

a = (K + 5), b = 13, c = -6k

Using b² - 4ac,

$\displaystyle (13)² - 4(k + 5)(-6k) $

$\displaystyle = 169 + 24k² + 120k $

$\displaystyle = 24k^2 + 120k + 169 $

Using the formula of finding the roots of a quadratic equation,

$\displaystyle \frac{-120 \pm \sqrt{(120)^2 - 4(24)(169)}}{2(24)} $

$\displaystyle \frac{-120 \pm \sqrt{-1824}}{2(24)} $

After this, I don't know what to do. I know that this way is wrong so please correct me on the steps.