#3x^2-5x>8#

Subtract #8# from both sides:

#3x^2-5x-8>0#

We now solve the equation:

#3x^2-5x-8=0#

This will give us the boundary values of #x#.

Factoring:

#(3x-8)(x+1)=0=> x=-1 and x= 8/3#

**METHOD 1**

From:

#3x^2-5x-8>0#

Notice that the coefficient of #x^2>0#, this means the parabola is in this form:

#uuu#

For values greater than #0# we will be above the #x# axis.

Since the roots to the equation are on the #x# axis we can see that for positive #y# values we will be to the left of #x=-1# and to the right of #x=8/3#.

Since this is a #># and not a #>= # inequality #x=-1# and #x=8/3# are not included points. So the solution in interval notation we be a union of intervals:

#(-oo,-1)uu(8/3,oo)#

**METHOD 2**

Using the factors of #3x^2-5x-8>0# we found earlier, we can see that:

#(3x-8)(x+1)#

Will be positive i.e. #>0# if both brackets are positive or both brackets are negative. We can use a table to check this using the following inequalities:

#x<-1# , #color(white)(88)-1< x< 8/3# , #color(white)(88) 8/3 < x#

You can see from the table that if we assign #x# a value satisfying the inequality in each column. it is only for the inequalities #x> -1# and #8/3 < x# that the product of the brackets is positive.

So our solution in interval notation is:

#(-oo,-1)uu(8/3 , oo)color(white(88)# as before.

Graph

graph{y<3x^2-5x-8 [-32.48, 32.46, -16.24, 16.25]}