Answer
The solution set is{$\frac{3-\sqrt 17}{4} ,\frac{3+\sqrt 17}{4}$}.
Work Step by Step
$2x^2-3x=1$
$x^2-\frac{3}{2}x=\frac{1}{2}$ Divide each term by 2.
$x^2-\frac{3}{2}x+\frac{9}{16}=\frac{1}{2}+\frac{9}{16}$. Add $(\frac{3}{2}\cdot\frac{1}{2})^2=\frac{9}{16}$ to both sides to complete the square.
$(x-\frac{3}{4})^2=\frac{17}{16}$
$x-\frac{3}{4}=\pm\sqrt \frac{17}{16}$ Use the square root property and solve.
$x=\frac{3}{4}\pm\sqrt \frac{17}{16}$
$x=\frac{3}{4}\pm\frac{\sqrt 17}{4}$
$x=\frac{3}{4}+\frac{\sqrt 17}{4}$, $x=\frac{3}{4}-\frac{\sqrt 17}{4}$
$x=\frac{3+\sqrt 17}{4}$, $x=\frac{3-\sqrt 17}{4}$,
The solution set is {$\frac{3-\sqrt 17}{4} ,\frac{3+\sqrt 17}{4}$}.
