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

The solutions to the equation (3x+3)(x+2)=1 are x=(-9+sqrt(21))/6 ,x=(-9-sqrt(21))/6

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

The zeros of (3x+3)(x+2)=1 are x=(-9+sqrt(21))/6 ,x=(-9-sqrt(21))/6

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

x = \frac{- 9 + 21}{6}, x = \frac{- 9 - 21}{6}

Solution steps

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

Expand

(3 x + 3) (x + 2) : 3 x^{2} + 9 x + 6

3 x^{2} + 9 x + 6 = 1

Move

1

to the left side

3 x^{2} + 9 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

9^{2} - 4 \cdot 3 \cdot 5 = 21

x_{1, 2} = \frac{- 9 \pm 21}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 9 + 21}{2 \cdot 3}, x_{2} = \frac{- 9 - 21}{2 \cdot 3}

x = \frac{- 9 + 21}{2 \cdot 3} : \frac{- 9 + 21}{6}

x = \frac{- 9 - 21}{2 \cdot 3} : \frac{- 9 - 21}{6}

The solutions to the quadratic equation are:

x = \frac{- 9 + 21}{6}, x = \frac{- 9 - 21}{6}

25 b^{4} - 36 b^{2}

\frac{4 ^{5}}{4 ^{- 2}}

3 x + 6 = 27

2 (x + y)^{2} + 5 (x + y) + 2

\frac{x}{3} + 5 \geq 3

What are the solutions to the equation (3x+3)(x+2)=1 ?
The solutions to the equation (3x+3)(x+2)=1 are x=(-9+sqrt(21))/6 ,x=(-9-sqrt(21))/6
Find the zeros of (3x+3)(x+2)=1
The zeros of (3x+3)(x+2)=1 are x=(-9+sqrt(21))/6 ,x=(-9-sqrt(21))/6