What are the solutions to the equation (3x+1)^2-2x=29 ?

The solutions to the equation (3x+1)^2-2x=29 are x= 14/9 ,x=-2

Find the zeros of (3x+1)^2-2x=29

The zeros of (3x+1)^2-2x=29 are x= 14/9 ,x=-2

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(3 x + 1)^{2} - 2 x = 29

x = \frac{14}{9}, x = - 2

Decimal Notation

x = 1.55555 \dots, x = - 2

Solution steps

(3 x + 1)^{2} - 2 x = 29

Expand

(3 x + 1)^{2} - 2 x : 9 x^{2} + 4 x + 1

9 x^{2} + 4 x + 1 = 29

Move

29

to the left side

9 x^{2} + 4 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 9 ( - 28 )}{2 \cdot 9}

4^{2} - 4 \cdot 9 (- 28) = 32

x_{1, 2} = \frac{- 4 \pm 32}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- 4 + 32}{2 \cdot 9}, x_{2} = \frac{- 4 - 32}{2 \cdot 9}

x = \frac{- 4 + 32}{2 \cdot 9} : \frac{14}{9}

x = \frac{- 4 - 32}{2 \cdot 9} : - 2

The solutions to the quadratic equation are:

x = \frac{14}{9}, x = - 2

3 - ∣ 2 x + 8 ∣ \leq 1

\frac{3}{4} \cdot \frac{7}{8} - \frac{1}{2}

8 - x \geq 5

\frac{2 X + 3}{\times + 5} \geq 1

x^{3} - 7 x + 6 = x^{3} + 5 x^{2} - 2 x - 24

What are the solutions to the equation (3x+1)^2-2x=29 ?
The solutions to the equation (3x+1)^2-2x=29 are x= 14/9 ,x=-2
Find the zeros of (3x+1)^2-2x=29
The zeros of (3x+1)^2-2x=29 are x= 14/9 ,x=-2