103= 6/1+r +106/(1+r)^2
r=???
Can anyone help with this? thank you
Let's say $\displaystyle (1 + r) = x$
Then $\displaystyle 103 = \frac{6}{x} + \frac{106}{x^2}$
[Multiply straight through with $\displaystyle x^2$]
$\displaystyle 103x^2 - 6x - 106 = 0$
$\displaystyle x = 1.0440028316648209$
OR
$\displaystyle x = -0.9857504044803549$
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$\displaystyle 1 + r = 1.0440028316648209$
OR
$\displaystyle 1 + r = -0.9857504044803549$
You can solve for r from there.