What are the solutions to the equation (3x+2)(x+1)=1 ?

The solutions to the equation (3x+2)(x+1)=1 are x=(-5+sqrt(13))/6 ,x=(-5-sqrt(13))/6

Find the zeros of (3x+2)(x+1)=1

The zeros of (3x+2)(x+1)=1 are x=(-5+sqrt(13))/6 ,x=(-5-sqrt(13))/6

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

x = \frac{- 5 + 13}{6}, x = \frac{- 5 - 13}{6}

Decimal Notation

x = - 0.23240 \dots, x = - 1.43425 \dots

Solution steps

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

Expand

(3 x + 2) (x + 1) : 3 x^{2} + 5 x + 2

3 x^{2} + 5 x + 2 = 1

Move

1

to the left side

3 x^{2} + 5 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

5^{2} - 4 \cdot 3 \cdot 1 = 13

x_{1, 2} = \frac{- 5 \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 5 + 13}{2 \cdot 3}, x_{2} = \frac{- 5 - 13}{2 \cdot 3}

x = \frac{- 5 + 13}{2 \cdot 3} : \frac{- 5 + 13}{6}

x = \frac{- 5 - 13}{2 \cdot 3} : \frac{- 5 - 13}{6}

The solutions to the quadratic equation are:

x = \frac{- 5 + 13}{6}, x = \frac{- 5 - 13}{6}

y + x^{3} y^{4}

15 x^{2} - 34 x + 15

\frac{24 a ^{2} c ^{2}}{36 a c}

7 x^{2} = 49

b^{2} - 6 b + 9

What are the solutions to the equation (3x+2)(x+1)=1 ?
The solutions to the equation (3x+2)(x+1)=1 are x=(-5+sqrt(13))/6 ,x=(-5-sqrt(13))/6
Find the zeros of (3x+2)(x+1)=1
The zeros of (3x+2)(x+1)=1 are x=(-5+sqrt(13))/6 ,x=(-5-sqrt(13))/6