Answer
No real solutions.
Work Step by Step
Comparing $2x^{2}-6x+5=0$ with $ax^{2}+bx+c=0$, we get
$a=2$, $b=-6$ and $c=5$.
The quadratic formula is
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Substituting the values, we get
$x=\frac{-(-6) \pm \sqrt {(-6)^{2}-4(2)(5)}}{(2)(2)}=\frac{6\pm \sqrt {-4}}{4}$
As the square of a real number can never be negative, this equation has no real solutions.
