Answer
$\left\{\dfrac{3-\sqrt {17}}{4},\dfrac{3+\sqrt {17}}{4}\right\}$
Work Step by Step
Add $1$ to both sides.
$2x^2-3x-1+1=0+1$
Simplify.
$2x^2-3x=1$
Divide both sides by $2$.
$x^2-\dfrac{3}{2}x=\dfrac{1}{2}$
Complete the square by adding $\left(\dfrac{-\frac{3}{2}}{2}\right)^2=\dfrac{9}{16}$ to both sides.
$x^2-\dfrac{3}{2}x+\dfrac{9}{16}=\dfrac{1}{2}+\dfrac{9}{16}$
Simplify.
$x^2-\dfrac{3}{2}x+\dfrac{9}{16}=\dfrac{8+9}{16}$
$x^2-\dfrac{3}{2}x+\dfrac{9}{16}=\dfrac{17}{16}$
Factor the left side.
$\left(x-\dfrac{3}{4}\right)^2=\dfrac{17}{16}$
Take the square root of both sides.
$x-\dfrac{3}{4}=\pm \sqrt {\dfrac{17}{16}}$
$x-\dfrac{3}{4}=\pm\dfrac{\sqrt {17}}{4}$
Split the expression to obtain:
$x-\dfrac{3}{4}=\dfrac{\sqrt {17}}{4}\quad $ or $\quad x-\dfrac{3}{4}=-\dfrac{\sqrt {17}}{4}$
Add $\frac{3}{4}$ to both sides of both equations.
$x-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{\sqrt {17}}{4}+\dfrac{3}{4}\quad$ or $\quad x-\dfrac{3}{4}+\dfrac{3}{4}=-\dfrac{\sqrt {17}}{4}+\dfrac{3}{4}$
Simplify.
$x=\dfrac{3+\sqrt {17}}{4}\quad$ or $\quad x=\dfrac{3-\sqrt {17}}{4}$
Hence, the solution set is $\left\{\dfrac{3-\sqrt {17}}{4},\dfrac{3+\sqrt {17}}{4}\right\}$.
