Quadratic equation formula help / simplification

Question

I have this quadratic equation,

$ x^{2} + \frac{10}{3}x -\frac{80}{3} = 0 $

I use the quadratic formula to solve and simplify

$-10 \pm \frac{\sqrt{100-(4)(3)(-80)}}{6}$ = $ \frac{-10 \pm \sqrt{1060}}{6}$

my book says it should simplify to

$ \frac{1}{3} ( -5 \pm \sqrt{73}) $

but i cant get this simplification can anyone show me if they can? Thank you.

Answer 1

Your solution seems correct, indeed

$$x^{2} + \frac{10}{3}x -\frac{80}{3} = 0\iff3x^2+10x-80=0$$

$$ \frac{-10 \pm\sqrt{100-(4)(3)(-80)}}{6}=-\frac53\pm\frac{\sqrt{265}}{3}$$

Answer 2

$$x=\frac{-10 \pm \sqrt{100-(4)(3)(-80)}}{6} =\frac{-2\times 5 \pm \sqrt{4\times25-(4)(3)(-80)}}{6} =\frac{1}{6}\left(-2\times 5 \pm 2\sqrt{25+240} \right) =\frac{2}{6}\left(-5 \pm \sqrt{265} \right)=\frac{1}{3}\left(-5 \pm \sqrt{265} \right)$$