Please explain this?

In equation 2, they didn't do the subtraction:

#-16y+9y = -7y#

#12y^2 -16y+9y -12 -= 12y^2 -7y-12=0#

they are the same equation, but the second makes it easier to factorise the expression, by grouping.

#12y^2-7y -12 = 0#

the first step when factorising a quadratic expression by grouping is to multiply the first and last term together.

#12 * -12 = -144#

the next step is to find two numbers that add to make the second term, and multiply to make the product of the first and last term.

#-16 + 9 = -7#

#-16 * 9 = -144#

this is why #12y^2-7y -12 = 0# can later be written as #12y^2 + 16y - 9y - 12 = 0#.

see below for solution of #y:#

#12y^2 + 16y - 9y - 12 = 0#

#12y^2+16y = 4y(3y+4)#

#-9y - 12 = -3(3y+4)#

#12y^2 + 16y - 9y - 12 = 4y(3y+4) -3(3y+4)#

#3y+4# is a common factor, so it can be bracketed off.

#4y(3y+4) -3(3y+4) = (4y-3)(3y+4)#

in the equation to solve for #x#, #(4y-3)(3y+4) = 0#

#n * 0 = 0#

if either #4y-3# or #3y+4# is #0#, the product of both will be #0#.

#4y-3 = 0#

#4y = 0+3 = 3#

#y = 3/4#

#3y+4 =0#

#3y = -4#

#y = -4/3#

this gives the two values of #y:# #3/4# and #-4/3#.