What are the solutions to the equation (2x-4)^2=32 ?

The solutions to the equation (2x-4)^2=32 are x=2(1+sqrt(2)),x=2(1-sqrt(2))

Find the zeros of (2x-4)^2=32

The zeros of (2x-4)^2=32 are x=2(1+sqrt(2)),x=2(1-sqrt(2))

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(2 x - 4)^{2} = 32

x = 2 (1 + 2), x = 2 (1 - 2)

Decimal Notation

x = 4.82842 \dots, x = - 0.82842 \dots

Solution steps

(2 x - 4)^{2} = 32

For

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

the solutions are

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

Solve

2 x - 4 = 32 : x = 2 (1 + 2)

Solve

2 x - 4 = - 32 : x = 2 (1 - 2)

The solutions to the quadratic equation are:

x = 2 (1 + 2), x = 2 (1 - 2)

3 x^{2} - 4 x + 9 = 10 x + 3

x^{2} - 14 x = 6

y, 2 x^{2} - 3 y^{2} + 5 x y - 2 x + 3 y - 4 = 0

4 x^{2} + 7 mn x - 15 m^{2} n^{2}

x, z = x^{2} - 3 y^{2} + 5 x y

What are the solutions to the equation (2x-4)^2=32 ?
The solutions to the equation (2x-4)^2=32 are x=2(1+sqrt(2)),x=2(1-sqrt(2))
Find the zeros of (2x-4)^2=32
The zeros of (2x-4)^2=32 are x=2(1+sqrt(2)),x=2(1-sqrt(2))