#### Answer

$x=5~$ or $~x=12$

#### Work Step by Step

Let's consider a trinomial in this form: $~x^2+bx+c$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$
Then we can factor the trinomial as follows:
$~x^2+bx+c = (x+r)~(x+s)$
We can use the GCF to factor the left side of the given equation:
$3x^2-51x+180=0$
$(3)~(x^2-17x+60)=0$
To factor the left side of the equation, we need to find two numbers $r$ and $s$ such that $r+s = -17$ and $r\times s = 60$. We can see that $(-5)+(-12) = -17~$ and $(-5)\times (-12) = 60$
We can solve the equation as follows:
$3x^2-51x+180=0$
$(3)~(x^2-17x+60)=0$
$(3)~(x-5)(x-12)=0$
$x=5~$ or $~x=12$