Question 414488
solve the following inequalities.
12<6x^2+x

..
6x^2+x>12
First step is to set inequality=to zero, than solve for zeros of the quadratic equation.

6x^2+x-12=0
solve using following quadratic equation: with a=6,b=1.c=-12
..
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
..

x=(-1+-sqrt(1-4*6*-12)/2*6
 =(-1+-sqrt(289))/12
 =(-1+-17)/12=-18/12,16/12
x=-3/2
x=4/3
in factored form:
(x+3/2)(x-4/3)>0
Show these points,-3/2 and 4/3 on a number line.
By using test points you see that for x>4/3 the function is positive.
Similarly, you will see that for x<-3/2, the function is also positive
For -3/2 < x < 4/3, the function, however is negative which does not satisfy the  inequality.
The solution therefore as expressed in interval notation:
(-infinity,-3/2) U (4/3, infinity)