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
$(m+1/3)(m-2)=0$

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

How do you solve 3m ^ { 2} - 5m - 7= - 53m ^ { 2} - 5m - 7= - 5?