$P = x_{1} . x_{2} = \dfrac{c}{a}$
Demonstração:
$x_{1} . x_{2} = \left( \dfrac{-b + \sqrt[]{b^{2} - 4ac}}{2a} \right) . \left( \dfrac{-b - \sqrt[]{b^{2} - 4ac}}{2a} \right)$
$x_{1} . x_{2} = \dfrac{b^{2} + b.\sqrt[]{b^{2} - 4ac} -b.\sqrt[]{b^{2} - 4ac} - \left( \sqrt[]{b^{2}-4ac} \right)^{2} }{4a^{2}}$
$x_{1} . x_{2} = \dfrac{b^{2} - \left( b^{2} -4ac \right) }{4a^{2}}$
$x_{1} . x_{2} = \dfrac{b^{2} -b^{2} + 4ac}{4a^{2}}$
$x_{1} . x_{2} = \dfrac{4ac}{4a^{2}}$
$x_{1} . x_{2} = \dfrac{c}{a} \quad$ (conforme queríamos demonstrar)