Question 722631
{{{2m^2-m-15}}}

factoring

coefficient of middle term is -1

the product of first term and third term is 30

find two factors of 30 such that when you add them you get -1

Obviously they are -6and +5

{{{2m^2-6m+5m-15=0}}}


{{{2m(m-3)+5(m-3)=0}}}

{{{(2m+5)(m-3)=0}}}

either 2m+5 =0 OR m-3=0 for the equation to be true

if 2m+5 = 0
then 2m=-5
m=-5/2

if m-3 =0
then m=3

so m=-5/2 or 3

completing the squares method

{{{2m^2-m-15=0}}}

the first term co efficient has to be  1
divide by 2

{{{m^2-(m/2) -(15/2)=0}}}

{{{m^2-(m/2) =(15/2)}}}

find the third term on the left side so that it is a perfect square
third term = (1/2 * co-efficient of second term)^2

=(1/2*1/2)^2
= 1/16

{{{m^2-(m/2) =(15/2)}}}
add the third term to both sides

{{{m^2-(m/2)+(1/16) =(15/2)+(1/16)}}}

{{{(m-(1/4))^2= -121/16}}}

take the square root

m-(1/4) = +/- 11/4

so m= 3 or -
5/2