Solve y=P(1+r)^2,r

$$y = P ( 1 + r ) ^ { 2 } , r$$

Solve for P (complex solution)

$$\left\{\begin{matrix}P=\frac{y}{\left(r+1\right)^{2}}\text{, }&r\neq -1\\P\in \mathrm{C}\text{, }&y=0\text{ and }r=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+r\right)^{2}.

$$y=P\left(1+2r+r^{2}\right)$$

Use the distributive property to multiply P by 1+2r+r^{2}.

$$y=P+2Pr+Pr^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$P+2Pr+Pr^{2}=y$$

Combine all terms containing P.

$$\left(1+2r+r^{2}\right)P=y$$

The equation is in standard form.

$$\left(r^{2}+2r+1\right)P=y$$

Divide both sides by 1+2r+r^{2}.

$$\frac{\left(r^{2}+2r+1\right)P}{r^{2}+2r+1}=\frac{y}{r^{2}+2r+1}$$

Dividing by 1+2r+r^{2} undoes the multiplication by 1+2r+r^{2}.

$$P=\frac{y}{r^{2}+2r+1}$$

Divide y by 1+2r+r^{2}.

$$P=\frac{y}{\left(r+1\right)^{2}}$$

Solve for P

$$\left\{\begin{matrix}P=\frac{y}{\left(r+1\right)^{2}}\text{, }&r\neq -1\\P\in \mathrm{R}\text{, }&y=0\text{ and }r=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+r\right)^{2}.

$$y=P\left(1+2r+r^{2}\right)$$

Use the distributive property to multiply P by 1+2r+r^{2}.

$$y=P+2Pr+Pr^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$P+2Pr+Pr^{2}=y$$

Combine all terms containing P.

$$\left(1+2r+r^{2}\right)P=y$$

The equation is in standard form.

$$\left(r^{2}+2r+1\right)P=y$$

Divide both sides by 1+2r+r^{2}.

$$\frac{\left(r^{2}+2r+1\right)P}{r^{2}+2r+1}=\frac{y}{r^{2}+2r+1}$$

Dividing by 1+2r+r^{2} undoes the multiplication by 1+2r+r^{2}.

$$P=\frac{y}{r^{2}+2r+1}$$

Divide y by 1+2r+r^{2}.

$$P=\frac{y}{\left(r+1\right)^{2}}$$

Solve for r (complex solution)

$$\left\{\begin{matrix}r=-1+P^{-\frac{1}{2}}\sqrt{y}\text{; }r=-1-P^{-\frac{1}{2}}\sqrt{y}\text{, }&P\neq 0\\r\in \mathrm{C}\text{, }&y=0\text{ and }P=0\end{matrix}\right.$$

Solve for r

$$\left\{\begin{matrix}r=-\sqrt{\frac{y}{P}}-1\text{; }r=\sqrt{\frac{y}{P}}-1\text{, }&y\leq 0\text{ and }P<0\\r=-\sqrt{\frac{y}{P}}-1\text{; }r=\sqrt{\frac{y}{P}}-1\text{, }&y\geq 0\text{ and }P>0\\r\in \mathrm{R}\text{, }&y=0\text{ and }P=0\end{matrix}\right.$$