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(37))/6 ,x=(-7-sqrt(37))/6

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

The zeros of (3x+1)(x+2)=1 are x=(-7+sqrt(37))/6 ,x=(-7-sqrt(37))/6

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

x = \frac{- 7 + 37}{6}, x = \frac{- 7 - 37}{6}

Decimal Notation

x = - 0.15287 \dots, x = - 2.18046 \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 + 1 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 3 \cdot 1 = 37

x_{1, 2} = \frac{- 7 \pm 37}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 7 + 37}{2 \cdot 3}, x_{2} = \frac{- 7 - 37}{2 \cdot 3}

x = \frac{- 7 + 37}{2 \cdot 3} : \frac{- 7 + 37}{6}

x = \frac{- 7 - 37}{2 \cdot 3} : \frac{- 7 - 37}{6}

The solutions to the quadratic equation are:

x = \frac{- 7 + 37}{6}, x = \frac{- 7 - 37}{6}

y^{2} - x^{2}

\frac{x - 8}{2 x + 7} \geq 1

\frac{x + 1}{x - 3} \leq 0

\frac{( 2 x - 1 ) ( x - 4 ) ^{2}}{x - 9} < 0

(- \frac{7}{4})^{2}

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(37))/6 ,x=(-7-sqrt(37))/6
Find the zeros of (3x+1)(x+2)=1
The zeros of (3x+1)(x+2)=1 are x=(-7+sqrt(37))/6 ,x=(-7-sqrt(37))/6