## Algebra: A Combined Approach (4th Edition)

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