Answer
The solution set is\[\left\{ -3,\frac{1}{2} \right\}\].
Work Step by Step
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=2,\,\,b=5,\text{ and }c=-3\]
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{-5\pm \sqrt{{{\left( 5 \right)}^{2}}-4\times 2\times \left( -3 \right)}}{2\times 2} \\
& =\frac{-5\pm \sqrt{25+24}}{4} \\
& =\frac{-5\pm \sqrt{49}}{4} \\
& =\frac{-5\pm 7}{4}
\end{align}\]
Simplify further,
\[\begin{align}
& x=\frac{-5\pm 7}{4} \\
& =\frac{-5-7}{4},\frac{-5+7}{4} \\
& =-\frac{12}{4},\frac{2}{4} \\
& =-3,\frac{1}{2}
\end{align}\]
