Answer
The equation has no real solutions.
Work Step by Step
$-3x^{2}+6x=4\implies 3x^{2}-6x+4=0$
Comparing $3x^{2}-6x+4=0$ with $ax^{2}+bx+c=0$, we get
$a=3$, $b=-6$ and $c=4$.
The quadratic formula is
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Substituting the values, we get
$x=\frac{-(-6) \pm \sqrt {(-6)^{2}-4(3)(4)}}{(2)(3)}=\frac{6\pm \sqrt {-12}}{6}$
As the square of a real number can never be negative, this equation has no real solutions.
