What are the solutions to the equation 400=500(1-r)^2 ?

The solutions to the equation 400=500(1-r)^2 are r=(-2+sqrt(5))/(sqrt(5)),r=(2+sqrt(5))/(sqrt(5))

Find the zeros of 400=500(1-r)^2

The zeros of 400=500(1-r)^2 are r=(-2+sqrt(5))/(sqrt(5)),r=(2+sqrt(5))/(sqrt(5))

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400 = 500 (1 - r)^{2}

r = \frac{- 2 + 5}{5}, r = \frac{2 + 5}{5}

Solution steps

400 = 500 (1 - r)^{2}

Switch sides

500 (1 - r)^{2} = 400

Divide both sides by

500

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

For

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

the solutions are

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

Solve

1 - r = \frac{4}{5} : r = \frac{- 2 + 5}{5}

Solve

1 - r = - \frac{4}{5} : r = \frac{2 + 5}{5}

The solutions to the quadratic equation are:

r = \frac{- 2 + 5}{5}, r = \frac{2 + 5}{5}

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

\frac{5}{x} < 3

3 x^{3} + 2 x^{2} + 9 x + 6

6 \cdot 16

216 h^{2} - 150

What are the solutions to the equation 400=500(1-r)^2 ?
The solutions to the equation 400=500(1-r)^2 are r=(-2+sqrt(5))/(sqrt(5)),r=(2+sqrt(5))/(sqrt(5))
Find the zeros of 400=500(1-r)^2
The zeros of 400=500(1-r)^2 are r=(-2+sqrt(5))/(sqrt(5)),r=(2+sqrt(5))/(sqrt(5))