Answer :
ANSWER
[tex]x=\frac{1}{2}+\frac{\sqrt[]{3}}{2}\text{ and }x=\frac{1}{2}-\frac{\sqrt[]{3}}{2}[/tex]EXPLANATION
Given the equation,
[tex]2x^2-2x-1=0[/tex]We have to solve it for x.
We can solve this using the quadratic formula,
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex]In our equation a = 2, b = -2 and c = -1,
[tex]x=\frac{2\pm\sqrt[]{(-2)^2-4\cdot2\cdot(-1)}}{2\cdot2}[/tex][tex]x=\frac{2\pm\sqrt[]{4+8}}{4}[/tex][tex]x=\frac{2\pm\sqrt[]{12}}{4}=\frac{2\pm\sqrt[]{4\cdot3}}{4}=\frac{2\pm2\sqrt[]{3}}{4}[/tex]Distribute the denominator into the addition/subtraction,
[tex]x=\frac{2}{4}\pm\frac{2\sqrt[]{3}}{4}=\frac{1}{2}\pm\frac{\sqrt[]{3}}{2}[/tex]The values of x that are solution to this equation are,
[tex]x=\frac{1}{2}+\frac{\sqrt[]{3}}{2}\text{ and }x=\frac{1}{2}-\frac{\sqrt[]{3}}{2}[/tex]