Answer:
[tex]x= \frac{3+\sqrt{7}i}{4}}[/tex] or [tex]x=\frac{3-\sqrt{7}i}{4}}[/tex]
Step-by-step explanation:
[tex]2x^2-3x+2=0[/tex]
We complete the square as follows
Group the constant term on the right hand side
[tex]2x^2-3x=-2[/tex]
Divide through by 2,
[tex]x^2-\frac{3}{2}x=-1[/tex]
Add half the square of the coefficient of x to get;
[tex]x^2-\frac{3}{2}x+(-\frac{3}{4})^2=-1+(-\frac{3}{4})^2[/tex]
Simplify to get
[tex](x-\frac{3}{4})^2=-1+\frac{9}{16}[/tex]
[tex](x-\frac{3}{4})^2=-\frac{7}{16}[/tex]
Take the square root of both sides to get
[tex](x-\frac{3}{4})=\pm \sqrt{-\frac{7}{16}}[/tex]
Solve for x
[tex](x-\frac{3}{4})=\pm \frac{\sqrt{7}i}{4}}[/tex]
[tex]x=\frac{3}{4} \pm \frac{\sqrt{7}i}{4}}[/tex]
[tex]x= \frac{3+\sqrt{7}i}{4}}[/tex] or [tex]x=\frac{3-\sqrt{7}i}{4}}[/tex]
