Answer
The equation has no real solution.
Work Step by Step
Comparing $x^{2}-6x+10=0$ with $ax^{2}+bx+c$, we see that
$a=1, b=-6$ and $c=10$.
$b^{2}-4ac=(-6)^{2}-4(1)(10)=36-40=-4$.
The discriminant is less than $0$. So, the equation has no real solution.
