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

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

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

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

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(k + 2)^{2} = 28

k = 27 - 2, k = - 27 - 2

Solution steps

(k + 2)^{2} = 28

For

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

the solutions are

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

Solve

k + 2 = 28 : k = 27 - 2

Solve

k + 2 = - 28 : k = - 27 - 2

The solutions to the quadratic equation are:

k = 27 - 2, k = - 27 - 2

\frac{x - 1}{x + 1} \leq 2

w - 2 + 2 w = 6 + 5 w

(x + 2) (x - 1) \leq 0

46

5 x^{2} = 6 - 13 x

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