Answer
$x=2$ and $x=-\dfrac{1}{2}$ satisfy the given conditions.
Work Step by Step
$y=2x^{2}-3x$ and $y=2$
Substitute $y$ by $2$:
$2=2x^{2}-3x$
Take all terms to the right side of the equation:
$0=2x^{2}-3x-2$
Rearrange:
$2x^{2}-3x-2=0$
Solve by factoring:
$(2x+1)(x-2)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$2x+1=0$
$2x=-1$
$x=-\dfrac{1}{2}$
$x-2=0$
$x=2$
$x=2$ and $x=-\dfrac{1}{2}$ satisfy the given conditions.