What are the solutions to the equation (3x-4)(x+1)=-2 ?

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

Find the zeros of (3x-4)(x+1)=-2

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

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(3 x - 4) (x + 1) = - 2

x = 1, x = - \frac{2}{3}

Decimal Notation

x = 1, x = - 0.66666 \dots

Solution steps

(3 x - 4) (x + 1) = - 2

Expand

(3 x - 4) (x + 1) : 3 x^{2} - x - 4

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

Move

2

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 3 (- 2) = 5

x_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 3}, x_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 3}

x = \frac{- ( - 1 ) + 5}{2 \cdot 3} : 1

x = \frac{- ( - 1 ) - 5}{2 \cdot 3} : - \frac{2}{3}

The solutions to the quadratic equation are:

x = 1, x = - \frac{2}{3}

30 x^{2} - 47 x + 14 = 0

2 5^{2} = 7^{2} + x^{2}

x^{2} + 9 x = 4

d (7 d - 6) + 4 = 0

2 x^{2} - 7 x - 13 = 0

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