#### Answer

$x=10$
$x={\dfrac{11}{3}}$

#### Work Step by Step

$x=\dfrac{-b±{\sqrt {b^{2}-4ac}}}{2a}$ and $3x^{2}-41x+110=0$.Therefore, $a=3$,$b=-41$ and $c=110$:
Substitute the values:
$x=\dfrac{41±{\sqrt {-41^{2}-4(3)(110)}}}{2(3)}$
=$\dfrac{41±{\sqrt {1681-1320}}}{6}$
=$\dfrac{41± {\sqrt {361}}}{6}$
=$\dfrac{41±19}{6}$
Separate the equation into plus and minus equations:
$x=\dfrac{41+19}{6}=10$
$x=\dfrac{41-19}{6}={\dfrac{11}{3}}$