What are the solutions to the equation -6x-2=(3x+1)^2 ?

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

Find the zeros of -6x-2=(3x+1)^2

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

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

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

Decimal Notation

x = - 0.33333 \dots, x = - 1

Solution steps

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

Expand

(3 x + 1)^{2} : 9 x^{2} + 6 x + 1

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

Switch sides

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

Move

2

to the left side

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

Move

6 x

to the left side

9 x^{2} + 12 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 9 \cdot 3}{2 \cdot 9}

1 2^{2} - 4 \cdot 9 \cdot 3 = 6

x_{1, 2} = \frac{- 12 \pm 6}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 12 + 6}{2 \cdot 9}, x_{2} = \frac{- 12 - 6}{2 \cdot 9}

x = \frac{- 12 + 6}{2 \cdot 9} : - \frac{1}{3}

x = \frac{- 12 - 6}{2 \cdot 9} : - 1

The solutions to the quadratic equation are:

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

40 - 9 x^{2} + 4 x + 4 - 12 - 4 x + 30 x^{2} + 30 x = 0

\frac{1}{3} a^{2} + a = 0

x^{2} = 5 (5 + 15)

\frac{1}{5} a^{2} - \frac{3}{5} a = 0

2 x^{2} + 6 x + 1

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