What are the solutions to the equation 6(3x+1)^2-9=3 ?

The solutions to the equation 6(3x+1)^2-9=3 are x=(sqrt(2)-1)/3 ,x=(-sqrt(2)-1)/3

Find the zeros of 6(3x+1)^2-9=3

The zeros of 6(3x+1)^2-9=3 are x=(sqrt(2)-1)/3 ,x=(-sqrt(2)-1)/3

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

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} - 9 = 3

Move

9

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}

(\frac{4 x ^{4} y ^{2}}{6 x y})^{2}

42 x + 18 = - 24 + 98 x

∣ 4 x + 3 ∣ < 2

16 x^{7} \cdot 48 x^{4}

3 x + 7 = 52 - 2 x

What are the solutions to the equation 6(3x+1)^2-9=3 ?
The solutions to the equation 6(3x+1)^2-9=3 are x=(sqrt(2)-1)/3 ,x=(-sqrt(2)-1)/3
Find the zeros of 6(3x+1)^2-9=3
The zeros of 6(3x+1)^2-9=3 are x=(sqrt(2)-1)/3 ,x=(-sqrt(2)-1)/3