Answer
$x=\left\{ -1,\dfrac{5}{2} \right\}$
Work Step by Step
Using the completing the square method, the solutions of the given quadratic equation, $
2x^2-3x-5=0
,$ is
\begin{array}{l}\require{cancel}
\dfrac{1}{2}\cdot(2x^2-3x-5)=(0)\cdot\dfrac{1}{2}
\\\\
x^2-\dfrac{3}{2}x-\dfrac{5}{2}=0
\\\\
x^2-\dfrac{3}{2}x=\dfrac{5}{2}
\\\\
x^2-\dfrac{3}{2}x+\left( -\dfrac{3}{2\cdot2}\right)^2=\dfrac{5}{2}+\left( -\dfrac{3}{2\cdot2}\right)^2
\\\\
x^2-\dfrac{3}{2}x+\dfrac{9}{16}=\dfrac{5}{2}+\dfrac{9}{16}
\\\\
\left( x-\dfrac{3}{4} \right)^2=\dfrac{40}{16}+\dfrac{9}{16}
\\\\
\left( x-\dfrac{3}{4} \right)^2=\dfrac{49}{16}
\\\\
x-\dfrac{3}{4}=\pm\sqrt{\dfrac{49}{16}}
\\\\
x-\dfrac{3}{4}=\pm\dfrac{7}{4}
\\\\
x=\dfrac{3}{4}\pm\dfrac{7}{4}
\\\\
x=\dfrac{3}{4}-\dfrac{7}{4}
\\\\
x=-\dfrac{4}{4}
\\\\
x=-1
,\\\\\text{OR}\\\\
x=\dfrac{3}{4}+\dfrac{7}{4}
\\\\
x=\dfrac{10}{4}
\\\\
x=\dfrac{5}{2}
.\end{array}
Hence, $
x=\left\{ -1,\dfrac{5}{2} \right\}
.$
