What are the solutions to the equation (2x+4)^2=5 ?

The solutions to the equation (2x+4)^2=5 are x=(sqrt(5)-4)/2 ,x=(-sqrt(5)-4)/2

Find the zeros of (2x+4)^2=5

The zeros of (2x+4)^2=5 are x=(sqrt(5)-4)/2 ,x=(-sqrt(5)-4)/2

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(2 x + 4)^{2} = 5

x = \frac{5 - 4}{2}, x = \frac{- 5 - 4}{2}

Decimal Notation

x = - 0.88196 \dots, x = - 3.11803 \dots

Solution steps

(2 x + 4)^{2} = 5

For

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

the solutions are

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

Solve

2 x + 4 = 5 : x = \frac{5 - 4}{2}

Solve

2 x + 4 = - 5 : x = \frac{- 5 - 4}{2}

The solutions to the quadratic equation are:

x = \frac{5 - 4}{2}, x = \frac{- 5 - 4}{2}

49 x^{2} + 1 = 25

5 x^{2} - 30 x + 25 = 0

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

2 x (5 x - 3) = 0

y, h = 16 - 3 y^{2}

What are the solutions to the equation (2x+4)^2=5 ?
The solutions to the equation (2x+4)^2=5 are x=(sqrt(5)-4)/2 ,x=(-sqrt(5)-4)/2
Find the zeros of (2x+4)^2=5
The zeros of (2x+4)^2=5 are x=(sqrt(5)-4)/2 ,x=(-sqrt(5)-4)/2