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

The solutions to the equation (3x+2)^2=(2x+1)(2x-1)-2 are x=-1,x=-7/5

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

The zeros of (3x+2)^2=(2x+1)(2x-1)-2 are x=-1,x=-7/5

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

x = - 1, x = - \frac{7}{5}

Decimal Notation

x = - 1, x = - 1.4

Solution steps

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

Expand

(3 x + 2)^{2} : 9 x^{2} + 12 x + 4

Expand

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

9 x^{2} + 12 x + 4 = 4 x^{2} - 3

Move

3

to the left side

9 x^{2} + 12 x + 7 = 4 x^{2}

Move

4 x^{2}

to the left side

5 x^{2} + 12 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 5 \cdot 7}{2 \cdot 5}

1 2^{2} - 4 \cdot 5 \cdot 7 = 2

x_{1, 2} = \frac{- 12 \pm 2}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 12 + 2}{2 \cdot 5}, x_{2} = \frac{- 12 - 2}{2 \cdot 5}

x = \frac{- 12 + 2}{2 \cdot 5} : - 1

x = \frac{- 12 - 2}{2 \cdot 5} : - \frac{7}{5}

The solutions to the quadratic equation are:

x = - 1, x = - \frac{7}{5}

x, k x^{2} - k x + 1 - 2 x = 0

x, \frac{x}{y} = \frac{- 3 x ^{2} + 6 x - 3}{2 y}

x^{2} + 3 x + 8 = 6

24 - 9 x^{2} = 2, 4 - 18 = 6

x, x^{2} + 4 y^{2} + z^{2} - 4 x - 8 y + 8 z + 15 = 0

What are the solutions to the equation (3x+2)^2=(2x+1)(2x-1)-2 ?
The solutions to the equation (3x+2)^2=(2x+1)(2x-1)-2 are x=-1,x=-7/5
Find the zeros of (3x+2)^2=(2x+1)(2x-1)-2
The zeros of (3x+2)^2=(2x+1)(2x-1)-2 are x=-1,x=-7/5