What are the solutions to the equation 100(1+I)^2=200 ?

The solutions to the equation 100(1+I)^2=200 are I=sqrt(2)-1,I=-sqrt(2)-1

Find the zeros of 100(1+I)^2=200

The zeros of 100(1+I)^2=200 are I=sqrt(2)-1,I=-sqrt(2)-1

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100 (1 + I)^{2} = 200

I = 2 - 1, I = - 2 - 1

Decimal Notation

I = 0.41421 \dots, I = - 2.41421 \dots

Solution steps

100 (1 + I)^{2} = 200

Divide both sides by

100

(1 + I)^{2} = 2

For

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

the solutions are

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

Solve

1 + I = 2 : I = 2 - 1

Solve

1 + I = - 2 : I = - 2 - 1

The solutions to the quadratic equation are:

I = 2 - 1, I = - 2 - 1

\frac{8 x}{6} = \frac{72}{54}

9 n^{2} + 12 n + 4

1 - x^{2} \leq 0

3 (5 - f) \geq 21 - 3 f

\frac{( - 3 ) ^{0}}{( - 3 ) ^{2}}

What are the solutions to the equation 100(1+I)^2=200 ?
The solutions to the equation 100(1+I)^2=200 are I=sqrt(2)-1,I=-sqrt(2)-1
Find the zeros of 100(1+I)^2=200
The zeros of 100(1+I)^2=200 are I=sqrt(2)-1,I=-sqrt(2)-1