Question

PLEASE HELP
solve 2v^2+4v=6 by factoring.

Answer 1

<=>v²+2v-3=0

<=>v²-v+3v-3=0

<=>v(v-1)+3(v-1)=0

<=>(v-1)(v+3)=0

<=>*v-1=0

v=1

*v+3=0

v=-3

Answer 2

v = 1 or -3

Explanation:

2v^2 + 4v = 6

take 2 as common

2(v^2 + 2v = 3)

v^2 + 2v = 3

Now,

v^2 + 2v - 3 = 0

v^2 - v + 3v - 3 = 0

v(v - 1) + 3(v - 1) = 0

(v - 1) (v + 3) = 0

v = 1 or -3