What are the solutions to the equation (r+4/5)^2= 1/25 ?

The solutions to the equation (r+4/5)^2= 1/25 are r=-3/5 ,r=-1

Find the zeros of (r+4/5)^2= 1/25

The zeros of (r+4/5)^2= 1/25 are r=-3/5 ,r=-1

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(r + \frac{4}{5})^{2} = \frac{1}{25}

r = - \frac{3}{5}, r = - 1

Decimal Notation

r = - 0.6, r = - 1

Solution steps

(r + \frac{4}{5})^{2} = \frac{1}{25}

For

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

the solutions are

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

Solve

r + \frac{4}{5} = \frac{1}{25} : r = - \frac{3}{5}

Solve

r + \frac{4}{5} = - \frac{1}{25} : r = - 1

The solutions to the quadratic equation are:

r = - \frac{3}{5}, r = - 1

- 16 t^{2} + 32 t + 28 = 0

12 x^{2} + 6 x = 0

6 x^{2} = - 7 x + 3

- 8 - 3 x + 2 x^{2} = 3

3 x^{2} = 33

What are the solutions to the equation (r+4/5)^2= 1/25 ?
The solutions to the equation (r+4/5)^2= 1/25 are r=-3/5 ,r=-1
Find the zeros of (r+4/5)^2= 1/25
The zeros of (r+4/5)^2= 1/25 are r=-3/5 ,r=-1