$$-1=(0.55)\cdot[1+(y+1)^2]^{\frac{3}{2}}$$
I got stuck with this expression. I have l some difficulty in leanding with some algebraic manipulation. What should I do to solve this equation??
I tried to answer but without results. I thought in taking square roots of booth sides, but this will lead me to the complex numbers.
Deep breath and be careful.
$-1= (0.55) \cdot[1 + (y+1)^2]^{\frac 32}$
$-1 = \frac {55}{100}\cdot[1 + (y+1)^2]^{\frac 32}$
$-\frac {100}{55} = -\frac {20}{11} = [1 + (y+1)^2]^{\frac 32}$
$(-\frac {20}{11})^{\frac 23} = 1 + (y+1)^2$
$(-\frac {20}{11})^{\frac 23} -1 = (y+1)^2$
$[(-\frac {20}{11})^{\frac 23} -1]^{\frac 12} =y +1$
$[(-\frac {20}{11})^{\frac 23} -1]^{\frac 12} -1 = y$
Now there are three values for $(-\frac {20}{11})^{\frac 23}$ (they are $\sqrt[3]{\frac {20}{11}}^2,\sqrt[3]{\frac {20}{11}}^2e^{\frac 23\pi i},\sqrt[3]{\frac {20}{11}}^2e^{\frac 43\pi}$).
ANd so there are six values for $[(-\frac {20}{11})^{\frac 23} -1]^{\frac 12} -1$