Answer :
ANSWER
[tex]x=1+\frac{\sqrt[]{2}}{2}i;x=1-\frac{\sqrt[]{2}}{2}i[/tex]EXPLANATION
We want to solve the given equation using the quadratic formula:
[tex]2x^{2}-4x+3=0[/tex]The quadratic formula is:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where a is the coefficient of x², b is the coefficient of x and c is the constant term.
Therefore, we have that:
[tex]\begin{gathered} x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(2)(3)}}{2(2)} \\ x=\frac{4\pm\sqrt[]{16-24}}{4} \\ x=\frac{4\pm\sqrt[]{(-8)}}{4} \\ x=\frac{4\pm2\sqrt[]{2}i}{4} \\ \Rightarrow x=1+\frac{\sqrt[]{2}}{2}i;x=1-\frac{\sqrt[]{2}}{2}i \end{gathered}[/tex]That is the solution of the equation.