Answer
Add $\frac{25}{16}$ to both sides of the equation to complete the square.
The solutions are $\frac{5}{4}±\frac{\sqrt31}{4}i$, and the solution set is {$\frac{5}{4}+\frac{\sqrt31}{4}i, \frac{5}{4}-\frac{\sqrt31}{4}i$}.
Work Step by Step
$2x^2 - 5x + 7 = 0$
Divide the whole equation by $2$. Thus, the equation becomes:
$x^2 - \frac{5}{2}x +\frac{7}{2} = 0$
Subtract $\frac{7}{2}$ from both sides to isolate the binomial $x^2 - \frac{5}{2}x$.
$x^2 - \frac{5}{2}x +\frac{7}{2} - \frac{7}{2} = 0 - \frac{7}{2}$
$x^2 - \frac{5}{2}x = - \frac{7}{2}$
The coefficient of the x-term is $- \frac{5}{2}$. Half of $- \frac{5}{2}$ is $- \frac{5}{4}$, and $(- \frac{5}{4})^2 = \frac{25}{16}$.
Thus, add $\frac{25}{16}$ to both sides of the equation to complete the square.
$x^2 - \frac{5}{2}x +\frac{25}{16}= - \frac{7}{2} +\frac{25}{16}$
$x^2 - \frac{5}{2}x +\frac{25}{16}= - \frac{31}{16}$
$(x-\frac{5}{4})^2 = - \frac{31}{16}$
$x-\frac{5}{4} = \sqrt {- \frac{31}{16}}$ or $x-\frac{5}{4} = -\sqrt {- \frac{31}{16}}$
$x = \frac{5}{4} +\sqrt {- \frac{31}{16}}$ or $x =\frac{5}{4} -\sqrt {- \frac{31}{16}}$
$x = \frac{5}{4}+\sqrt \frac{31}{16}\sqrt {-1}$ or $x = \frac{5}{4} -\sqrt \frac{31}{16}\sqrt {-1}$
$x =\frac{5}{4}+\frac{\sqrt{31}}{4}i$ or $x = \frac{5}{4}-\frac{\sqrt{31}}{4}i$
The solutions are $\frac{5}{4}±\frac{\sqrt{31}}{4}i$, and the solution set is {$\frac{5}{4}+\frac{\sqrt{31}}{4}i, \frac{5}{4}-\frac{\sqrt{31}}{4}i$}.
