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

The solutions to the equation (3x-2)(x+1)=2 are x=1,x=-4/3

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

The zeros of (3x-2)(x+1)=2 are x=1,x=-4/3

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

x = 1, x = - \frac{4}{3}

Solution steps

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

Expand

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

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

Move

2

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 1 + 7}{2 \cdot 3} : 1

x = \frac{- 1 - 7}{2 \cdot 3} : - \frac{4}{3}

The solutions to the quadratic equation are:

x = 1, x = - \frac{4}{3}

\frac{31.827 - 30.656}{30.656 - 30.114}

5 + x - 14 = x - 7

∣ x - 12 ∣ = \frac{x}{3}

x^{2} - 7 x = 18

∣ 2 x - 1 ∣ = 8 x

What are the solutions to the equation (3x-2)(x+1)=2 ?
The solutions to the equation (3x-2)(x+1)=2 are x=1,x=-4/3
Find the zeros of (3x-2)(x+1)=2
The zeros of (3x-2)(x+1)=2 are x=1,x=-4/3