Answer
$x=\frac{3}{2}$ and $x=\frac{2}{3}$
Work Step by Step
$6x^{2}-13x=-6\implies 6x^{2}-13x+6=0$
Comparing $6x^{2}-13x+6=0$ with $ax^{2}+bx+c=0$, we get
$a=6$, $b=-13$ and $c=6$.
The quadratic formula is
$x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$
Substituting the values, we get
$x=\frac{-(-13) \pm \sqrt {(-13)^{2}-4(6)(6)}}{(2)(6)}=\frac{13\pm 5}{12}$
The solutions of the equation are $x=\frac{13+5}{12}=\frac{18}{12}=\frac{3}{2}$ and $x=\frac{13-5}{12}=\frac{8}{12}=\frac{2}{3}$
