Respuesta :
Given the equation:
[tex]x^2-2x+3=0[/tex]Let's find the roots of the equation.
To find the roots of the equation, let's solve for x using the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{4a}[/tex]Use the standard equation to find the values of a, b, and c:
[tex]\begin{gathered} ax^2+bx+c=0 \\ \\ x^2-2x+3=0 \end{gathered}[/tex]a = 1
b = -2
c = 3
Input the values into the quadratic equation and solve for x:
[tex]\begin{gathered} x=\frac{-(-2)\pm\sqrt{-2^2-4(1)(3)}}{2(1)} \\ \\ x=\frac{2\pm\sqrt{4-12}}{2} \\ \\ x=\frac{2\pm\sqrt{-8}}{2} \end{gathered}[/tex]Solving further:
[tex]\begin{gathered} x=\frac{2\pm\sqrt{(8)(-1)}}{2} \\ \\ \text{ Where:} \\ \sqrt{-1}=i \\ \\ x=\frac{2\pm i\sqrt{8}}{2} \\ \\ x=\frac{2\pm i\sqrt{(2)(4)}}{2} \\ \\ x=\frac{2\pm i\sqrt{(2)(2)^2}}{2} \end{gathered}[/tex]Therefore, we have:
[tex]\begin{gathered} x=\frac{2\pm2i\sqrt{2}}{2} \\ \\ \text{ Divide all terms by 2:} \\ x=\frac{2}{2}\pm\frac{2i\sqrt{2}}{2} \\ \\ x=1\pm i\sqrt{2} \end{gathered}[/tex]The roots of the equation:
[tex]x=1\pm i\sqrt{2}[/tex]ANSWER:
[tex]1\pm i\sqrt{2}[/tex]