Answer
$\left\{\dfrac{1}{2},1\right\}$
Work Step by Step
In the form $x^2+bx=c,$ the given equation, $
2x^2-3x=-1
,$ is equivalent to
\begin{align*}
\dfrac{2x^2-3x}{2}&=-\dfrac{1}{2}
\\\\
x^2-\dfrac{3}{2}x&=-\dfrac{1}{2}
.\end{align*}
Completing the square of the left-hand expression equation above is equivalent to
\begin{align*}
x^2-\dfrac{3}{2}x+\left(\dfrac{-3/2}{2}\right)^2&=-\dfrac{1}{2}+\left(\dfrac{-3/2}{2}\right)^2
&(\text{add }\left(\dfrac{b}{2}\right)^2)
\\\\
x^2-\dfrac{3}{2}x+\dfrac{9}{16}&=-\dfrac{1}{2}+\dfrac{9}{16}
\\\\
\left(x-\dfrac{3}{4}\right)^2&=-\dfrac{8}{16}+\dfrac{9}{16}
\\\\
\left(x-\dfrac{3}{4}\right)^2&=\dfrac{1}{16}
.\end{align*}
Taking the square root of both sides (Square Root Property), the equation above is equivalent to
\begin{align*}
x-\dfrac{3}{4}&=\pm\sqrt{\dfrac{1}{16}}
\\\\
x-\dfrac{3}{4}&=\pm\sqrt{\left(\dfrac{1}{4}\right)^2}
\\\\
x-\dfrac{3}{4}&=\pm\dfrac{1}{4}
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
x&=\dfrac{3}{4}\pm\dfrac{1}{4}
\end{align*}\begin{array}{c|c}
x=\dfrac{3}{4}-\dfrac{1}{4} & x=\dfrac{3}{4}+\dfrac{1}{4}
\\\\
x=\dfrac{2}{4} & x=\dfrac{4}{4}
\\\\
x=\dfrac{1}{2} & x=1
.\end{array}
Hence, the solution set of $
2x^2-3x=-1
$ is $
\left\{\dfrac{1}{2},1\right\}
$.
