What are the solutions to the equation (k+7)^2=289 ?

The solutions to the equation (k+7)^2=289 are k=10,k=-24

Find the zeros of (k+7)^2=289

The zeros of (k+7)^2=289 are k=10,k=-24

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(k + 7)^{2} = 289

k = 10, k = - 24

Solution steps

(k + 7)^{2} = 289

For

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

the solutions are

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

Solve

k + 7 = 289 : k = 10

Solve

k + 7 = - 289 : k = - 24

The solutions to the quadratic equation are:

k = 10, k = - 24

x^{2} + 2 = 6 x + 4

x (x - 2) = 24

15 x^{2} - 75 x + 90 = 0

- x^{2} + 5 x + 9 = 0

x^{2} - 20 x + 36 = 0

What are the solutions to the equation (k+7)^2=289 ?
The solutions to the equation (k+7)^2=289 are k=10,k=-24
Find the zeros of (k+7)^2=289
The zeros of (k+7)^2=289 are k=10,k=-24