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

The solutions to the equation (3x+1)^2=x+2 are x=(-5+sqrt(61))/(18),x=(-5-sqrt(61))/(18)

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

The zeros of (3x+1)^2=x+2 are x=(-5+sqrt(61))/(18),x=(-5-sqrt(61))/(18)

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

x = \frac{- 5 + 61}{18}, x = \frac{- 5 - 61}{18}

Decimal Notation

x = 0.15612 \dots, x = - 0.71168 \dots

Solution steps

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

Expand

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

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

Move

2

to the left side

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

Move

x

to the left side

9 x^{2} + 5 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 9 ( - 1 )}{2 \cdot 9}

5^{2} - 4 \cdot 9 (- 1) = 61

x_{1, 2} = \frac{- 5 \pm 61}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 5 + 61}{2 \cdot 9}, x_{2} = \frac{- 5 - 61}{2 \cdot 9}

x = \frac{- 5 + 61}{2 \cdot 9} : \frac{- 5 + 61}{18}

x = \frac{- 5 - 61}{2 \cdot 9} : \frac{- 5 - 61}{18}

The solutions to the quadratic equation are:

x = \frac{- 5 + 61}{18}, x = \frac{- 5 - 61}{18}

x, y^{2} = - 6 x^{2} + c

6 x^{2} - 42 x + 24 = 0

x^{2} - 24 x - 81 = 0

3 x^{2} - 2 x - 5 = 0

(2 x - 3)^{2} + 6 = 0

What are the solutions to the equation (3x+1)^2=x+2 ?
The solutions to the equation (3x+1)^2=x+2 are x=(-5+sqrt(61))/(18),x=(-5-sqrt(61))/(18)
Find the zeros of (3x+1)^2=x+2
The zeros of (3x+1)^2=x+2 are x=(-5+sqrt(61))/(18),x=(-5-sqrt(61))/(18)