What are the solutions to the equation (x+4)^2=121 ?

The solutions to the equation (x+4)^2=121 are x=7,x=-15

Find the zeros of (x+4)^2=121

The zeros of (x+4)^2=121 are x=7,x=-15

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(x + 4)^{2} = 121

x = 7, x = - 15

Solution steps

(x + 4)^{2} = 121

For

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

the solutions are

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

Solve

x + 4 = 121 : x = 7

Solve

x + 4 = - 121 : x = - 15

The solutions to the quadratic equation are:

x = 7, x = - 15

x \leq - 3 and x \geq - 2

\frac{1}{2} (0)

- x^{2} + 8 x - 15 < 0

(x + y)^{2} (x - y)^{2}

\frac{6}{4}

What are the solutions to the equation (x+4)^2=121 ?
The solutions to the equation (x+4)^2=121 are x=7,x=-15
Find the zeros of (x+4)^2=121
The zeros of (x+4)^2=121 are x=7,x=-15