Question 148747
Dear Ellen,


Traditional method:
{{{2x^2-3x-5}}}=0
First factor it out:


Start with
(2x )(x )


Then notice that everything after {{{2x^2}}} is negative so
(2x- )(x+ )


Then insert two factors of 5 (the only two would be 5 and 1):
(2x-5)(x+1)


Set each set of parentheses equal to 0:
(2x-5)=0
(x+1)=0


Solve for x in both:
2x-5=0
2x=5
x=5/2


x+1=0
x=-1


So x=5/2; -1
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Indian method:
{{{2x^2-3x-5}}}=0

Multiply the whole thing by 4(2) or 8:

{{{16x^2-24x-40}}}=0

Factor
{{{(4x-10)(4x+4)=0}}}

Set each expression equal to zero and solve
(4x-10)=0
4x=10
x=10/4
Reduce
x=5/2


(4x+4)=0
4x=-4
x=-1


So x=5/2; -1


Sincerely,
     Alani