#### Answer

$x=-2.363$ is a reasonably accurate solution of $f(x)=3x^4+4x^3-6x^2-2x-12$ in the interval $[-3,-2]$
$x=1.496$ is a reasonably accurate solution of $f(x)=3x^4+4x^3-6x^2-2x-12$ in the interval $[1,2]$

#### Work Step by Step

We are given $3x^4+4x^3-6x^2-2x-12=0$
which is also $f(x)=3x^4+4x^3-6x^2-2x-12$
$f'(x)=12x^3+12x^2-12x-2$
Since $f(-3)$ and $f(-2)$ have opposite signs, there is a solution for the equation in the interval $[-3,-2]$
As an initial guess, let $c_1=-3$
$c_2$ can be found as follows:
$c_2=c_1-\frac{f(c_1)}{f'(c_1)} \approx -2.5879$
In the same way,
$c_3=-2.403$
$c_4=-2.3648$
$c_5=-2.363$
$x=-2.363$ is a reasonably accurate solution of $f(x)=3x^4+4x^3-6x^2-2x-12$ in the interval $[-3,-2]$
Since $f(1)$ and $f(2)$ have opposite signs, there is a solution for the equation in the interval $[1,2]$
As an initial guess, let $c_1=1$
$c_2$ can be found as follows:
$c_2=c_1-\frac{f(c_1)}{f'(c_1)} = 2.3$
In the same way,
$c_3=1.831$
$c_4=1.581$
$c_5=1.5032$
$c_6=1.496$
$c_7=1.496$
$x=1.496$ is a reasonably accurate solution of $f(x)=3x^4+4x^3-6x^2-2x-12$ in the interval $[1,2]$