What are the solutions to the equation (3x-4)^2-1=7 ?

The solutions to the equation (3x-4)^2-1=7 are x=(2sqrt(2)+4)/3 ,x=(-2sqrt(2)+4)/3

Find the zeros of (3x-4)^2-1=7

The zeros of (3x-4)^2-1=7 are x=(2sqrt(2)+4)/3 ,x=(-2sqrt(2)+4)/3

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(3 x - 4)^{2} - 1 = 7

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

Decimal Notation

x = 2.27614 \dots, x = 0.39052 \dots

Solution steps

(3 x - 4)^{2} - 1 = 7

Move

1

to the right side

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

For

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

the solutions are

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

Solve

3 x - 4 = 8 : x = \frac{2 2 + 4}{3}

Solve

3 x - 4 = - 8 : x = \frac{- 2 2 + 4}{3}

The solutions to the quadratic equation are:

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

3 (5 x - 7)^{2} - 23 = - 17

0 = 6 x^{2}

7 v^{2} - 42 = - 35 v

3 x^{2} - 7 x - 6 = 0

2 x^{2} - x - 29 = 15

What are the solutions to the equation (3x-4)^2-1=7 ?
The solutions to the equation (3x-4)^2-1=7 are x=(2sqrt(2)+4)/3 ,x=(-2sqrt(2)+4)/3
Find the zeros of (3x-4)^2-1=7
The zeros of (3x-4)^2-1=7 are x=(2sqrt(2)+4)/3 ,x=(-2sqrt(2)+4)/3