Answer
$\displaystyle \frac{5}{2}$ and $-2$
Work Step by Step
Set the $RHS$ to 0 by subtracting 10 from both sides
$2k^{2}-k=10\qquad.../-10$
$2k^{2}-k-10=0$
To factor the trinomial, we search for
factors of $2\times(-10)=-20, $ whose sum is $-1:\qquad $
We find $-5$ and +4.
$2k^{2}-k-10$=$2k^{2}+4k-5k-10$
... and factor in pairs:
$2k(k+2)-5(k+2)=(2k-5)(k+2)$
Our equation becomes
$(2k-5)(k+2)=0$
By the zero product principle,
$2k-5=0$ or $k+2=0$
$k=\displaystyle \frac{5}{2}$ or $k=-2$
The solutions are $\displaystyle \frac{5}{2}$ and $-2$.
