Given: #33x^2-x-14=0#

Since there is an #x^2# we know there will be two factors for this equation so we can start setting up the factor brackets:

#(x ...)(x ...) = 0#

We then need to find factors of both #33# and #14# that will add or subtract to leave a value of #-1# which is the multiplier of the middle or #x# term.

Factors of #33# are #33*1; 11*3;# and that's it.

Factors of #14# are #14*1; 7*2;# and that's it.

Addition or subtraction of factors #33*1 and 14*1# will not give us #-1#.

Addition or subtraction of factors #11*3 and 7*2# may give us #-1#.

Place the factors into the brackets:

#(11x ...7)(3x ...2) = 0#

We can see that #11*2=22 and 7*3=21# which is needed to give the #-1# we need, if the first is negative and the second is positive.

We also know that one factor will contain a positive integer and the other will contain a negative integer because the #14# is negative.

Inserting the signs: #(11x+7)(3x-2) = 0#

Now we can solve for the two factors of #x#:

#11x+7=0; 11x=-7; x=-7/11#

#3x-2=0; 3x=2; x=2/3#

To check, substitute answers into the #given# equation:

#33x^2-x-14=0#

#33(2/3)^2-2/3-14=0#

#33(4/9)-2/3-14=0#

#cancel(33)(4/cancel(9))^2-14 2/3=0#

#44/3-14 2/3=0#