(3x+2)(x+1)=1...solve by any method?

Answer 1

3x62+3x+2x+2=1
you do the rest

Answer 2

Ok. Lets see if it's easy.
If its easy the two binomials are either both 1 or both -1
The second is 1 if x=0, but this makes the first 2
The second is -1 if x= -2 but this makes the first -4
So it's not easy.

The hard way:
Multiply out the left, then set equal to 0
3x^2 + 5x + 2 = 1
3x^2 + 5x + 1 = 0
solving by the quadratic formula gives
x = -.232
or
x = -1.434

Answer 3

(3x + 2)(x + 1) = 1

3x^2 + 5x + 2 = 1 {multiply the binomials}

3x^2 + 5x + 1 = 0

Use quadratic formula

x = [- 5 +- sqrt(25 - 12)]/6

x = [- 5 +- sqrt(13)]/6

Answer 4

Multiply through (FOIL)
3x^2+5x+2=1

To factor or use quadratic formula, must have 0 on right side so subtract 1 from both sides to get:

3x^2+5x+1=0

Then use quadratic formula to get (-5 ± √(25-12))/6

That is [-5 ± √(13)]/6

simplify to solve for x

Answer 5

Use the FOIL methode F as in First, Out, I as in In and L as in Last.
(3x+2)(x+1)=1
F= 3x^2
O= 3x
I=2x
L=2
so it's 3x^2+3x+2x+2 which becomes
3x^2+5x+2=1
you subtract one to get 3x^2+5x+1=0
the quadratic formula is x=-b plus or minus square root of b^2-4ac divided by 2a
a=3, b=5 and c=1 plus those values in
x=-5 plus or minus sqre rt of 25-4(3)(1)
then x= -5 plus or minus sqre rt of 25-12 then
x=-5 plus or minus sqre rt of 13 divided by 6
FINAL ANSWER:
x=-5 plus or minus sqre rt of 13 divided by 6

Answer 6

x=-1/3 or 0

Answer 7

3 x ² + 5 x + 1 = 0
x = [- 5 ± √(25 - 12) ] / 6
x = [- 5 ± √ 13 ] / 6

Answer 8

3x^2 + 5x - 1

Answer 9

3x + 2 = 1

3x = -1

x = -1/3
or
x + 1 = 1

x = 0

ANSWER: x = -1/3 or 0

Answer 10

3x^2+5x+3 = 0

x^2 + 5/3x +1 = 0

x1 = -5/6 + i*root { 4 - (5/3)^2 }/2 , i = root(-1)
x2 = -5/6 - i*root { 4 - (5/3)^2 }/2 , i = root(-1)