What are the solutions to the equation 6(3x+1)^2-4=8 ?

The solutions to the equation 6(3x+1)^2-4=8 are x=(sqrt(2)-1)/3 ,x=(-sqrt(2)-1)/3

Find the zeros of 6(3x+1)^2-4=8

The zeros of 6(3x+1)^2-4=8 are x=(sqrt(2)-1)/3 ,x=(-sqrt(2)-1)/3

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6 (3 x + 1)^{2} - 4 = 8

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

Decimal Notation

x = 0.13807 \dots, x = - 0.80473 \dots

Solution steps

6 (3 x + 1)^{2} - 4 = 8

Move

4

to the right side

6 (3 x + 1)^{2} = 12

Divide both sides by

6

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

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

3 x + 1 = 2 : x = \frac{2 - 1}{3}

Solve

3 x + 1 = - 2 : x = \frac{- 2 - 1}{3}

The solutions to the quadratic equation are:

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

x - 3 \leq 7

\frac{k - 5}{2} = 2

(2 x + 5) (2 x - 5) = 11

30 x^{3} + 22 x^{2} + 4 x

t^{2} - 5 t - 36

What are the solutions to the equation 6(3x+1)^2-4=8 ?
The solutions to the equation 6(3x+1)^2-4=8 are x=(sqrt(2)-1)/3 ,x=(-sqrt(2)-1)/3
Find the zeros of 6(3x+1)^2-4=8
The zeros of 6(3x+1)^2-4=8 are x=(sqrt(2)-1)/3 ,x=(-sqrt(2)-1)/3