Answer
$5 +\sqrt{5}\approx 7.236$
$5 -\sqrt{5}\approx 2.764$
Work Step by Step
Set the $RHS$ to 0 by adding 20 to both sides
$k^{2}-10k=-20\qquad.../+20$
$k^{2}-10k+20=0$
To factor the trinomial we search for
integer factors of $20$ whose sum is $-10$ ...
... none
We can always use the quadratic formula:
(For $ax^{2}+bx+c=0,\ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ )
$k=\displaystyle \frac{-(-10)\pm\sqrt{(-10)^{2}-4(1)(20)}}{2(1)}$
$=\displaystyle \frac{10\pm\sqrt{100-80}}{2}=\frac{10\pm\sqrt{20}}{2}$
$=\displaystyle \frac{10\pm\sqrt{4\times 5}}{2}=\frac{10\pm 2\sqrt{5}}{2}$
$=\displaystyle \frac{2(5\pm\sqrt{5})}{2}$
$k=5\pm\sqrt{5}$
Solutions:
$5 +\sqrt{5}\approx 7.236$ and
$5 -\sqrt{5}\approx 2.764$.
