#3x^2 + 11x -4=0#
#ax^2 +bx +c=0#
Multiply #ac# to get #-12#
find factors the multiply to get #-12# and add to get the coefficient of the middle term #+11#
Because we want #-12#, one factor is negative and the other is positive. Because we want the sum to be #+11#, the factor with greater absolute value is the positive factor:
List:
#-1xx12# sum #-1+12 = 11# STOP!, that's the one we want.
Now write the quadratic, replacing the middle term #11x# withe the two numbers we just found: #-x+12x#
#3x^2-x+12x-4 = 0# Factor by grouping:
#(3x^2-x)+(12x-4) = 0#
#x(3x-1)+4(3x-1) = 0#
#(x+4)(3x-1)=0#
#x+4=0# or #3x-1 = 0#
#x= -4# or #x= 1/3#
The solutions are #-4# and #1/3#
