#### Answer

$x=\frac{1}{3}~~$ or $~~x=\frac{1}{2}$

#### Work Step by Step

Let's consider a trinomial in this form: $~ax^2+bx+c$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = a\times c$
Then the next step is to rewrite the trinomial as follows:
$~x^2+bx+c = ax^2+rx+sx+c$
We can rewrite the given equation:
$6x^2 = 5x-1$
$6x^2 - 5x +1 = 0$
To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = -5~$ and $~r\times s = 6$. We can see that $(-2)+(-3) = -5~$ and $(-2)\times (-3) = 6$
We can solve the equation as follows:
$6x^2 = 5x-1$
$6x^2 - 5x +1 = 0$
$6x^2 - 3x - 2x +1 = 0$
$(6x^2 - 3x)+(-2x +1) = 0$
$(3x)(2x - 1)+(-1)(2x - 1) = 0$
$(3x-1)~(2x-1) = 0$
$3x-1 = 0~~$ or $~~2x-1 = 0$
$x=\frac{1}{3}~~$ or $~~x=\frac{1}{2}$