#### Answer

$x=-\frac{2}{3}~~$ or $~~x=4$

#### 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 write the trinomial as follows:
$~x^2+bx+c = ax^2+rx+sx+c$
We can rewrite the given equation:
$3x^2 = 10x+8$
$3x^2 - 10x - 8 = 0$
To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = -10~$ and $~r\times s = -24$. We can see that $(-12)+(2) = -10~$ and $(-12)\times (2) = -24$
We can solve the equation as follows:
$3x^2 = 10x+8$
$3x^2 - 10x - 8 = 0$
$3x^2 - 12x+2x - 8 = 0$
$(3x^2 - 12x)+(2x - 8) = 0$
$(3x)(x - 4)+(2)(x - 4) = 0$
$(3x+2)~(x-4) = 0$
$3x+2 = 0~~$ or $~~x-4 = 0$
$x=-\frac{2}{3}~~$ or $~~x=4$