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

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

Find the zeros of (3x+4)^2=2

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

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(3 x + 4)^{2} = 2

x = \frac{2 - 4}{3}, x = \frac{- 2 - 4}{3}

Decimal Notation

x = - 0.86192 \dots, x = - 1.80473 \dots

Solution steps

(3 x + 4)^{2} = 2

For

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

the solutions are

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

Solve

3 x + 4 = 2 : x = \frac{2 - 4}{3}

Solve

3 x + 4 = - 2 : x = \frac{- 2 - 4}{3}

The solutions to the quadratic equation are:

x = \frac{2 - 4}{3}, x = \frac{- 2 - 4}{3}

2 x - 3 x^{2} = 5

20 (t^{2} + 16) = 320 t

(2 t - 3)^{2} - 8 (2 t - 3) + 7 = 0

8 = 3 t^{2} - 5 t + 1

2 r^{2} + 6 r = - 12 - 2 r

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