What are the solutions to the equation 3244.8=3000(1+r/1)^2 ?

The solutions to the equation 3244.8=3000(1+r/1)^2 are r=sqrt(1.0816)-1,r=-sqrt(1.0816)-1

Find the zeros of 3244.8=3000(1+r/1)^2

The zeros of 3244.8=3000(1+r/1)^2 are r=sqrt(1.0816)-1,r=-sqrt(1.0816)-1

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3244.8 = 3000 (1 + \frac{r}{1})^{2}

r = 1.0816 - 1, r = - 1.0816 - 1

Decimal Notation

r = 0.04, r = - 2.04

Solution steps

3244.8 = 3000 (1 + \frac{r}{1})^{2}

Switch sides

3000 (1 + \frac{r}{1})^{2} = 3244.8

Divide both sides by

3000

(1 + \frac{r}{1})^{2} = 1.0816

For

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

the solutions are

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

Solve

1 + \frac{r}{1} = 1.0816 : r = 1.0816 - 1

Solve

1 + \frac{r}{1} = - 1.0816 : r = - 1.0816 - 1

The solutions to the quadratic equation are:

r = 1.0816 - 1, r = - 1.0816 - 1

p, x^{2} - 2 p x + p^{2} + p - 6 = 0

3 x^{2} - 15 = 66

x^{2} - 6 x + 5 = - 3

x^{2} - 10 x - 15 = 0

12 x^{2} - 2 x = 18

What are the solutions to the equation 3244.8=3000(1+r/1)^2 ?
The solutions to the equation 3244.8=3000(1+r/1)^2 are r=sqrt(1.0816)-1,r=-sqrt(1.0816)-1
Find the zeros of 3244.8=3000(1+r/1)^2
The zeros of 3244.8=3000(1+r/1)^2 are r=sqrt(1.0816)-1,r=-sqrt(1.0816)-1