Answer
$x_{1}=3+ \sqrt{7}$ and $x_{2}=3- \sqrt{7}$
Work Step by Step
Given $u^2-6u+2=0$
$a=1, \ b=-6, \ c=2$
Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-(-6)\pm \sqrt{(-6)^2-4\times 1\times 2}}{2\times 1} = \dfrac{6\pm \sqrt{36-8}}{2} = \dfrac{6\pm \sqrt{28}}{2} = \dfrac{6\pm \sqrt{4\times 7}}{2} = \dfrac{6\pm 2\sqrt{7}}{2} = 3\pm \sqrt{7}$
Therefore the solutions are $x_{1}=3+ \sqrt{7}$ and $x_{2}=3- \sqrt{7}$
