Answer
$x=2+\dfrac{\sqrt{6}}{2}$ and $x=2-\dfrac{\sqrt{6}}{2}$ are not part of the domain of this expression.
Work Step by Step
$\dfrac{7}{2x^{2}-8x+5}$
The values of $x$ that are not included in the domain of this expression are the ones that make its denominator equal to $0$.
Set the denominator equal to $0$:
$2x^{2}-8x+5=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=2$, $b=-8$ and $c=5$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-8)\pm\sqrt{(-8)^{2}-4(2)(5)}}{2(2)}=\dfrac{8\pm\sqrt{64-40}}{4}=...$
$...=\dfrac{8\pm\sqrt{24}}{4}=\dfrac{8\pm2\sqrt{6}}{4}=2\pm\dfrac{\sqrt{6}}{2}$
$x=2+\dfrac{\sqrt{6}}{2}$ and $x=2-\dfrac{\sqrt{6}}{2}$ are not part of the domain of this expression.
