Question 179905
{{{x^2-x-2=0}}}
You can factor this quadratic equation and solve for the zeros. 
{{{x^2-x-2=0}}}
{{{(x-2)(x+1)=0}}}
First solution:
{{{x-2=0}}}
{{{highlight( x=2)}}}
Second solution:
{{{x+1=0}}}
{{{highlight( x=-1)}}}
.
.
.
Here's a graph of the function to verify the solutions.
{{{drawing( 300, 300, -5, 5, -5, 5,circle( 2, 0, .2 ), circle(-1,0,.2),graph( 300, 300, -5, 5, -5, 5, x^2-x-2)) }}}
.
.
.
You could also use the quadratic formula,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
which is used to solve the general quadratic equation,
{{{ax^2+bx+c=0}}}
In your case,
{{{a=1}}}
{{{b=-1}}}
{{{c=-2}}}
Then,
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-(-1) +- sqrt( (-1)^2-4*1*(-2) ))/(2*1) }}}
{{{x = (1 +- sqrt( 1+8 ))/(2) }}}
{{{x = (1 +- sqrt( 9))/(2) }}}
{{{x = (1 +- 3)/(2) }}}
First solution:
{{{x = (1 + 3)/(2) }}}
{{{x = (4)/(2) }}}
{{{highlight(x=2)}}}
Second solution:
{{{x = (1 - 3)/(2) }}}
{{{x = (-2)/(2) }}}
{{{highlight(x=-1)}}}