What are the solutions to the equation (x+5)^2=121 ?

The solutions to the equation (x+5)^2=121 are x=6,x=-16

Find the zeros of (x+5)^2=121

The zeros of (x+5)^2=121 are x=6,x=-16

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(x + 5)^{2} = 121

x = 6, x = - 16

Solution steps

(x + 5)^{2} = 121

For

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

the solutions are

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

Solve

x + 5 = 121 : x = 6

Solve

x + 5 = - 121 : x = - 16

The solutions to the quadratic equation are:

x = 6, x = - 16

\frac{x}{2} \cdot \frac{x}{4} = \frac{3}{4}

4 x^{2} - 2 x - 2 = 0

x^{2} + 5 x + 1

5 k^{2} - 9 k + 18 = 4 k^{2}

s^{2} + 4 s + 13

What are the solutions to the equation (x+5)^2=121 ?
The solutions to the equation (x+5)^2=121 are x=6,x=-16
Find the zeros of (x+5)^2=121
The zeros of (x+5)^2=121 are x=6,x=-16