What are the solutions to the equation (x+4)^2=81 ?

The solutions to the equation (x+4)^2=81 are x=5,x=-13

Find the zeros of (x+4)^2=81

The zeros of (x+4)^2=81 are x=5,x=-13

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(x + 4)^{2} = 81

x = 5, x = - 13

Solution steps

(x + 4)^{2} = 81

For

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

the solutions are

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

Solve

x + 4 = 81 : x = 5

Solve

x + 4 = - 81 : x = - 13

The solutions to the quadratic equation are:

x = 5, x = - 13

16 \cdot 9

7 \cdot \frac{1}{9}

250 x^{3} - 128 x^{6}

(2 + 8)^{2}

\frac{5}{2} - x = 3 x

What are the solutions to the equation (x+4)^2=81 ?
The solutions to the equation (x+4)^2=81 are x=5,x=-13
Find the zeros of (x+4)^2=81
The zeros of (x+4)^2=81 are x=5,x=-13