Answer
$x=\frac{3-\sqrt {57}}{6},\frac{3+\sqrt {57}}{6}$
Work Step by Step
Plug the values $a$, $b$, and $c$ into the quadratic formula and simplify all the values.
$3x^2-3x=4=0$
$a=3,b=-3,c=-4$
$x=\frac{-b±\sqrt {b^2-4ac}}{2a}$
$x=\frac{-(-3)±\sqrt {(-3)^2-4(3)(-4)}}{2(3)}$
$x=\frac{3±\sqrt {9+48}}{6}$
$x=\frac{3±\sqrt {57}}{6}$
$x=\frac{3-\sqrt {57}}{6},\frac{3+\sqrt {57}}{6}$
