What are the solutions to the equation (x+2)^2=28 ?

The solutions to the equation (x+2)^2=28 are x=2sqrt(7)-2,x=-2sqrt(7)-2

Find the zeros of (x+2)^2=28

The zeros of (x+2)^2=28 are x=2sqrt(7)-2,x=-2sqrt(7)-2

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(x + 2)^{2} = 28

x = 27 - 2, x = - 27 - 2

Solution steps

(x + 2)^{2} = 28

For

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

the solutions are

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

Solve

x + 2 = 28 : x = 27 - 2

Solve

x + 2 = - 28 : x = - 27 - 2

The solutions to the quadratic equation are:

x = 27 - 2, x = - 27 - 2

- 5 < x \leq 4

- 12 r + 4 = 100

x^{4}

y^{2} = 12 y - 36

6 x + 3 x - 4 = 14

What are the solutions to the equation (x+2)^2=28 ?
The solutions to the equation (x+2)^2=28 are x=2sqrt(7)-2,x=-2sqrt(7)-2
Find the zeros of (x+2)^2=28
The zeros of (x+2)^2=28 are x=2sqrt(7)-2,x=-2sqrt(7)-2