How do you solve #\frac { 3x ^ { 2} + 10x + 8} { - x - 2} = - 2#?

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Answer 1 · 2017-03-16T05:36:25

#x=-2/3#

You would get the equivalent equation:

#3x^2+10x+8=-2(-x-2)# and #color(red)(x!=-2)#

that's

#3x^2+10x+8=2x+4#

#3x^2+8x+4=0#

#x=(-8+-sqrt(64-48))/6#

#=(-8+-4)/6#

#color(red)(x_1=-2)#; #x_2=-2/3#

and you cannot admit the first solution