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

The solutions to the equation (3x+1)(x+2)=-1 are x=(-7+sqrt(13))/6 ,x=(-7-sqrt(13))/6

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

The zeros of (3x+1)(x+2)=-1 are x=(-7+sqrt(13))/6 ,x=(-7-sqrt(13))/6

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

x = \frac{- 7 + 13}{6}, x = \frac{- 7 - 13}{6}

Decimal Notation

x = - 0.56574 \dots, x = - 1.76759 \dots

Solution steps

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

Expand

(3 x + 1) (x + 2) : 3 x^{2} + 7 x + 2

3 x^{2} + 7 x + 2 = - 1

Move

1

to the left side

3 x^{2} + 7 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

7^{2} - 4 \cdot 3 \cdot 3 = 13

x_{1, 2} = \frac{- 7 \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 7 + 13}{2 \cdot 3}, x_{2} = \frac{- 7 - 13}{2 \cdot 3}

x = \frac{- 7 + 13}{2 \cdot 3} : \frac{- 7 + 13}{6}

x = \frac{- 7 - 13}{2 \cdot 3} : \frac{- 7 - 13}{6}

The solutions to the quadratic equation are:

x = \frac{- 7 + 13}{6}, x = \frac{- 7 - 13}{6}

2 x^{2} + 2 = 3 x

x^{2} + 5 x - 36

\frac{1}{1 - \frac{1}{4}}

∣ 10 - 4 x ∣ \leq 9

3 (y + 1)^{2} - 5 = 43

What are the solutions to the equation (3x+1)(x+2)=-1 ?
The solutions to the equation (3x+1)(x+2)=-1 are x=(-7+sqrt(13))/6 ,x=(-7-sqrt(13))/6
Find the zeros of (3x+1)(x+2)=-1
The zeros of (3x+1)(x+2)=-1 are x=(-7+sqrt(13))/6 ,x=(-7-sqrt(13))/6