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

The solutions to the equation (3x+3)(x+2)=2 are x=(-9+sqrt(33))/6 ,x=(-9-sqrt(33))/6

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

The zeros of (3x+3)(x+2)=2 are x=(-9+sqrt(33))/6 ,x=(-9-sqrt(33))/6

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

x = \frac{- 9 + 33}{6}, x = \frac{- 9 - 33}{6}

Decimal Notation

x = - 0.54257 \dots, x = - 2.45742 \dots

Solution steps

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

Expand

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

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

Move

2

to the left side

3 x^{2} + 9 x + 4 = 0

Solve with the quadratic formula

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

9^{2} - 4 \cdot 3 \cdot 4 = 33

x_{1, 2} = \frac{- 9 \pm 33}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 9 + 33}{2 \cdot 3}, x_{2} = \frac{- 9 - 33}{2 \cdot 3}

x = \frac{- 9 + 33}{2 \cdot 3} : \frac{- 9 + 33}{6}

x = \frac{- 9 - 33}{2 \cdot 3} : \frac{- 9 - 33}{6}

The solutions to the quadratic equation are:

x = \frac{- 9 + 33}{6}, x = \frac{- 9 - 33}{6}

\frac{2}{6} = \frac{w}{9}

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

- 2 \cdot 3^{- (- 1)}

0.5 + 1

h - 6 > 2

What are the solutions to the equation (3x+3)(x+2)=2 ?
The solutions to the equation (3x+3)(x+2)=2 are x=(-9+sqrt(33))/6 ,x=(-9-sqrt(33))/6
Find the zeros of (3x+3)(x+2)=2
The zeros of (3x+3)(x+2)=2 are x=(-9+sqrt(33))/6 ,x=(-9-sqrt(33))/6