Answer
$x=1$ and $x=-\frac{1}{2}$
Work Step by Step
Comparing $2x^{2}-x-1=0$ with $ax^{2}+bx+c=0$, we get
$a=2$, $b=-1$ and $c=-1$.
The quadratic formula is
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Substituting the values, we get
$x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(2)(-1)}}{(2)(2)}=\frac{1\pm 3}{4}$
The solutions are $x=\frac{1+3}{4}=1$ and $x=\frac{1-3}{4}=-\frac{1}{2}$
