How do you solve #3m ^ { 2} - 5m - 7= - 5#?

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Answer 1 · 2018-06-20T14:13:46

#m=-1/3# or #m=-2#

#3m^2-5m-7=-5#
#3m^2-5m-7+5=0#
#3m^2-5m-2=0#

Using either the criss cross method or the quadratic formula, you would get the answer
#(3m+1)(m-2)=0# so #m=-1/3# or #m=-2#

Using the quadratic formula,

#color (red) (m=(-b+-sqrt(b^2-4ac))/(2a)# where your general form is given by #am^2+bm+c=0#

#m=(5+-sqrt(25+24))/6#

#m=(5+-sqrt49)/6#

#m=(5+-7)/6#

#m=(5-7)/6# or #m=(5+7)/6#

#m=-1/3# or #m=2#

#(m-(-1/3))(m-2)=0#

Which can then be put into the following form
#(m+1/3)(m-2)=0#

#(3m+1)(m-2)=0#