What are the solutions to the equation 6(3x+1)^2-8=10 ?

The solutions to the equation 6(3x+1)^2-8=10 are x=(sqrt(3)-1)/3 ,x=(-sqrt(3)-1)/3

Find the zeros of 6(3x+1)^2-8=10

The zeros of 6(3x+1)^2-8=10 are x=(sqrt(3)-1)/3 ,x=(-sqrt(3)-1)/3

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6 (3 x + 1)^{2} - 8 = 10

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

Decimal Notation

x = 0.24401 \dots, x = - 0.91068 \dots

Solution steps

6 (3 x + 1)^{2} - 8 = 10

Move

8

to the right side

6 (3 x + 1)^{2} = 18

Divide both sides by

6

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

For

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

the solutions are

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

5 (x - 9)^{2} = 15

37 = 2 x^{2} - 40

- 2 (b + 5) = 4

\frac{33 !}{( 33 - 6 ) !}

(x - 1) \cdot (x - 2) \cdot (x + 2) \cdot (x + 3) - 5

What are the solutions to the equation 6(3x+1)^2-8=10 ?
The solutions to the equation 6(3x+1)^2-8=10 are x=(sqrt(3)-1)/3 ,x=(-sqrt(3)-1)/3
Find the zeros of 6(3x+1)^2-8=10
The zeros of 6(3x+1)^2-8=10 are x=(sqrt(3)-1)/3 ,x=(-sqrt(3)-1)/3