Answer
The solution set for the quadratic equation, $2{{x}^{2}}-6x+5=0$ is $\left\{ \frac{3}{2}\pm \frac{1}{2}i \right\}$.
Work Step by Step
Consider quadratic equation,
$2{{x}^{2}}-6x+5=0$
Compare the equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0\text{ , }\left( a\ne 0 \right)$.
Here, $a=2,\text{ }b=-6\text{ and }c=5$
Apply the quadratic formula
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substitute 2 for a, $-6$ for b and 5 for c.
$x=\frac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 2 \right)\left( 5 \right)}}{2\left( 2 \right)}$
Simplify the radical.
$\begin{align}
& x=\frac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 2 \right)\left( 5 \right)}}{2\left( 2 \right)} \\
& =\frac{6\pm \sqrt{36-40}}{4} \\
& =\frac{6\pm \sqrt{-4}}{4}
\end{align}$
As $i=\sqrt{-1}$
Therefore,
$\begin{align}
& x=\frac{6\pm \sqrt{-1}\sqrt{4}}{4} \\
& =\frac{6\pm 2i}{4} \\
& =\frac{2\left( 3\pm i \right)}{4} \\
& =\frac{3\pm i}{2}
\end{align}$
Convert it to standard form.
$\begin{align}
& x=\frac{3\pm i}{2} \\
& =\frac{3}{2}\pm \frac{1}{2}i
\end{align}$
The solutions of the quadratic are complex conjugates of each other.
Hence, the solution set for the quadratic equation $2{{x}^{2}}-6x+5=0$ is $\left\{ \frac{3}{2}\pm \frac{1}{2}i \right\}$.
