What are the solutions to the equation 3380=2000(1+r)^2 ?

The solutions to the equation 3380=2000(1+r)^2 are r= 3/10 ,r=-23/10

Find the zeros of 3380=2000(1+r)^2

The zeros of 3380=2000(1+r)^2 are r= 3/10 ,r=-23/10

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3380 = 2000 (1 + r)^{2}

r = \frac{3}{10}, r = - \frac{23}{10}

Decimal Notation

r = 0.3, r = - 2.3

Solution steps

3380 = 2000 (1 + r)^{2}

Switch sides

2000 (1 + r)^{2} = 3380

Divide both sides by

2000

(1 + r)^{2} = \frac{169}{100}

For

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

the solutions are

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

Solve

1 + r = \frac{169}{100} : r = \frac{3}{10}

Solve

1 + r = - \frac{169}{100} : r = - \frac{23}{10}

The solutions to the quadratic equation are:

r = \frac{3}{10}, r = - \frac{23}{10}

324

14 c^{2} d + 2 c d^{2}

2 x^{2} - 1 = 5 x

4 x^{2} + x - 3 = 0

\frac{x - 2}{3} + \frac{x - 2}{4} = \frac{7}{24}

What are the solutions to the equation 3380=2000(1+r)^2 ?
The solutions to the equation 3380=2000(1+r)^2 are r= 3/10 ,r=-23/10
Find the zeros of 3380=2000(1+r)^2
The zeros of 3380=2000(1+r)^2 are r= 3/10 ,r=-23/10