Question

after solving 6x^2 + 8x = 14
i was left with 3x^2+4x-7=0. after that i am unsure on how to finish it

Answer 1

3x^2-3x+7x-7=0
3x(x-1)+7(x-1)=0
(x-1)(3x+7)=0
x=1 or x= -7/3

Answer 2

does that help?

Answer 3

where did you get the 1 from?

Answer 4

did the 1 come from finding what number can be divided into both sets of numbers? 3/3=1and 7/7=1 ?

Answer 5

yea exactly

i took two sets of the terms in the equation

3x^2 - 3x

now take 3x common

3x(x-1)

same thing for the other two terms

Answer 6

i understand that part now, but where did the -7/3 come from?

Answer 7

(x-1)(3x+7)=0

so when we have

a*b=0 then it is a=0 or b=0

so we have

x-1=0

x=1 or

3x+7=0

3x=-7

x=-7/3

Answer 8

is that fine?

Answer 9

yeah so i just have to separate both sets of numbers and equal them to 0 and i should have my final answer?

Answer 10

x= 1 or -3/7

Answer 11

thank you i understand now. can you also explain how to solve (x+2)^2=1 i cant find any good examples

Answer 12

ok this can be solved in another way too

(x+2)^2=1

take square root on both the sides

(x+2)=1

x=-1 :D

Answer 13

what do you mean square root?

Answer 14

\[\sqrt{x}\]

Answer 15

oh! ok yeah i get it.

Answer 16

\[x^2 - x +6 = 0\]

i will take this example for you

Answer 17

can u try it?

Answer 18

is the answer x=3 or x=-2 ?

Answer 19

At a quick glance, I Beleive that is correct

Answer 20

wow im suprised i got it :D. reffering back to my second question, since x=-1 how do i find the other x? its either -3 or 4