How do you solve #2x^2 = 3x -2=0#?
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I think you meant #2x^2+3x-2 = 0# since #+# and #=# are on the same key.
#0 = 2x^2+3x-2 = (2x-1)(x+2)#
So #x = 1/2# or #x = -2#
Let #f(x) = 2x^2+3x-2#.
If #f(x)# has factors with integer coefficients, then they must be of the form #(2x+-1)(x+-2)# for some combination of #+-# signs.
To see this, first note that #2# only factors as #+-2 xx +- 1#. The coefficient of the #x^2# term is #2#, the constant term is #2# and the middle term is not divisible by #2#. So one of the factors is #(2x+-1)# and the other is #(x+-2)#
Furthermore, the sign of the constant term is #-#, so one of the factors has a #+# and the other a #-#. That leaves only two possibilities to try:
#(2x+1)(x-2) = 2x^2-3x-2#
and
#(2x-1)(x+2) = 2x^2+3x-2#
