How do you solve the equation: #(m-3)(m-4)=20#?

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Answer 1 · 2018-04-09T08:40:30

# m=8, or, m=-1#.

#(m−3)(m−4)=20#

#(m^2 -3m−4m +12)=20#

#m^2 -7m -8=0#

#m^2-8m +m -8=0#

#m(m-8) +1(m -8)=0#

#(m-8)(m +1)=0#

Therefore, either, #m-8=0# or #m+1=0#

That is, #m=8# or #m=-1#

Answer 2 · 2018-04-09T09:41:34

# 8 and -1#.

#(m-3)(m-4)=20=5xx4 rArr m-3=5 rArr m=8#.

#(m-3)(m-4)=20=(-4)(-5) rArr m-3=-4#,

#rArr m=-1#.

Since it is a quadr. eqn., it can not have more than #2# roots.