#### Answer

$x= (\frac{1+i\sqrt 13}{2}, \frac{1-i\sqrt 13}{2})$

#### Work Step by Step

Step-1 : Since the co-efficient of $x^2$ is not 1, we divide both sides of the equation of the co-efficient of $x^2$ which is 2
$x^{2}-x+\frac{7}{2} = 0$
Step-2: Subtract $\frac{7}{2}$ from both sides
$x^{2}-x = -\frac{7}{2} $
Step-3:
Co-efficient of $x=-1$
Half of co-efficient of $x = \frac{1}{2}\times-1=-\frac{1}{2}$
Square of $-\frac{1}{2} = (-\frac{1}{2})^{2} = \frac{1}{4}$
Step-4: We add $\frac{1}{4}$ to both sides of the equation
$x^{2}-x+\frac{1}{4} = -\frac{7}{2} + \frac{1}{4}$
Step-5: Factor the trinomial and simplify the right hand side
$(x-\frac{1}{2})^2= \frac{-14+1}{4}$
$(x-\frac{1}{2})^2= \frac{-13}{4}$
Step-6: Use the square root property and solve for $x$
$x-\frac{1}{2}= ±\sqrt \frac{-13}{4}$
$x-\frac{1}{2}= ±\frac{i\sqrt 13}{2}$
Step-7: Add $\frac{1}{2}$ on both sides of the equation
$x= \frac{1}{2} ±\frac{i\sqrt 13}{2}$
$x= \frac{1 ± i\sqrt 13}{2}$
Therefore the solution set is $(\frac{1+i\sqrt 13}{2}, \frac{1-i\sqrt 13}{2})$