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.

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