What are the solutions to the equation (2x-4)^2=23 ?

The solutions to the equation (2x-4)^2=23 are x=(sqrt(23)+4)/2 ,x=(-sqrt(23)+4)/2

Find the zeros of (2x-4)^2=23

The zeros of (2x-4)^2=23 are x=(sqrt(23)+4)/2 ,x=(-sqrt(23)+4)/2

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(2 x - 4)^{2} = 23

x = \frac{23 + 4}{2}, x = \frac{- 23 + 4}{2}

Decimal Notation

x = 4.39791 \dots, x = - 0.39791 \dots

Solution steps

(2 x - 4)^{2} = 23

For

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

the solutions are

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

Solve

2 x - 4 = 23 : x = \frac{23 + 4}{2}

Solve

2 x - 4 = - 23 : x = \frac{- 23 + 4}{2}

The solutions to the quadratic equation are:

x = \frac{23 + 4}{2}, x = \frac{- 23 + 4}{2}

(x^{\frac{5}{9}})^{\frac{2}{3}}

∣ 3 x - 1 ∣ < 2 x + 5

12 (x + 5) = 4 x

2 + 4 z \geq 10

4 n - 5 = 23

What are the solutions to the equation (2x-4)^2=23 ?
The solutions to the equation (2x-4)^2=23 are x=(sqrt(23)+4)/2 ,x=(-sqrt(23)+4)/2
Find the zeros of (2x-4)^2=23
The zeros of (2x-4)^2=23 are x=(sqrt(23)+4)/2 ,x=(-sqrt(23)+4)/2