Answer
$-1+\frac{-{{P}_{2}}\pm \sqrt{{{\left( {{P}_{2}} \right)}^{2}}+4{{P}_{1}}A}}{2{{P}_{1}}}$
Work Step by Step
$A={{P}_{1}}{{\left( 1+r \right)}^{2}}+{{P}_{2}}\left( 1+r \right)$
Subtract $A$ on both sides,
$\begin{align}
& A={{P}_{1}}{{\left( 1+r \right)}^{2}}+{{P}_{2}}\left( 1+r \right) \\
& A-A={{P}_{1}}{{\left( 1+r \right)}^{2}}+{{P}_{2}}\left( 1+r \right)-A \\
& 0={{P}_{1}}{{\left( 1+r \right)}^{2}}+{{P}_{2}}\left( 1+r \right)-A
\end{align}$
Let, $u=1+r$. So, the equation is,
$0={{P}_{1}}{{u}^{2}}+{{P}_{2}}u-A$
The equation is in the form $a{{x}^{2}}+bx+c$,
Substitute the values $a={{P}_{1}},b={{P}_{2}},c=-A$ in the quadratic equation.
$\begin{align}
& u=\frac{-{{P}_{2}}\pm \sqrt{{{\left( {{P}_{2}} \right)}^{2}}-4\left( {{P}_{1}} \right)\left( -A \right)}}{2{{P}_{1}}} \\
& =\frac{-{{P}_{2}}\pm \sqrt{{{\left( {{P}_{2}} \right)}^{2}}+4{{P}_{1}}A}}{2{{P}_{1}}}
\end{align}$
Replace $u$ with $1-r$ and solve for $r$,
$1+r=\frac{-{{P}_{2}}\pm \sqrt{{{\left( {{P}_{2}} \right)}^{2}}+4{{P}_{1}}A}}{2{{P}_{1}}}$
Subtract $-1$ from both sides of the equation,
$\begin{align}
& 1+r-\left( 1 \right)=\frac{-{{P}_{2}}\pm \sqrt{{{\left( {{P}_{2}} \right)}^{2}}+4{{P}_{1}}A}}{2{{P}_{1}}}-\left( 1 \right) \\
& r=-1+\frac{-{{P}_{2}}\pm \sqrt{{{\left( {{P}_{2}} \right)}^{2}}+4{{P}_{1}}A}}{2{{P}_{1}}}
\end{align}$
Thus, the value is $r=-1+\frac{-{{P}_{2}}\pm \sqrt{{{\left( {{P}_{2}} \right)}^{2}}+4{{P}_{1}}A}}{2{{P}_{1}}}$.
