Solve the following equations:
$\left( {x + 9} \right)\left( {x - 3} \right)\left( {x - 7} \right)\left( {x + 5} \right) = 385$

Hint: In this question first multiply two factors together and the remaining two together so that it converts into a quadratic equation then substitute the same part to any other variable and multiply later on to apply the quadratic formula, so use these concepts to get the solution of the question.

Given equation is
$\left( {x + 9} \right)\left( {x - 3} \right)\left( {x - 7} \right)\left( {x + 5} \right) = 385$
Now multiply (x+9) and (x-7) together and remaining two together
\[
  \left[ {\left( {x + 9} \right)\left( {x - 7} \right)} \right]\left[ {\left( {x - 3} \right)\left( {x + 5} \right)} \right] = 385 \\
  \left[ {{x^2} + 9x - 7x - 63} \right]\left[ {{x^2} - 3x + 5x - 15} \right] = 385 \\
  \left[ {{x^2} + 2x - 63} \right]\left[ {{x^2} + 2x - 15} \right] = 385 \\
\]
Let, $\left( {{x^2} + 2x} \right) = t...........\left( 1 \right)$
So, substitute this value in the above equation we have
\[
  \left[ {t - 63} \right]\left[ {t - 15} \right] = 385 \\
   \Rightarrow {t^2} - 15t - 63t + 945 = 385 \\
   \Rightarrow {t^2} - 78t + 560 = 0 \\
\]
Now factorize the above equation we have,
\[
   \Rightarrow {t^2} - 8t - 70t + 560 = 0 \\
   \Rightarrow t\left( {t - 8} \right) - 70\left( {t - 8} \right) = 0 \\
   \Rightarrow \left( {t - 8} \right)\left( {t - 70} \right) = 0 \\
   \Rightarrow \left( {t - 8} \right) = 0{\text{ & }}\left( {t - 70} \right) = 0 \\
   \Rightarrow t = 8{\text{ & }}t = 70 \\
\]
Now from equation (1)
\[
  \left( {{x^2} + 2x} \right) = t \\
   \Rightarrow \left( {{x^2} + 2x} \right) = 8{\text{ & }}\left( {{x^2} + 2x} \right) = 70 \\
   \Rightarrow {x^2} + 2x - 8 = 0..........\left( 2 \right){\text{ & }}{x^2} + 2x - 70 = 0........\left( 3 \right) \\
\]
Now first solve equation (2) by factorization method we have,
\[
   \Rightarrow {x^2} + 2x - 8 = 0 \\
   \Rightarrow {x^2} + 4x - 2x - 8 = 0 \\
   \Rightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \\
   \Rightarrow \left( {x + 4} \right) = 0{\text{ & }}\left( {x - 2} \right) = 0 \\
   \Rightarrow x = - 4{\text{ & }}x = 2 \\
\]
Now solve equation (3)
\[{x^2} + 2x - 70 = 0\]
This is a quadratic equation so apply quadratic formula
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4\left( 1 \right)\left( { - 70} \right)} }}{2} = \dfrac{{ - 2 \pm \sqrt {284} }}{2} = \dfrac{{ - 2 \pm 2\sqrt {71} }}{2} = - 1 \pm \sqrt {71} $
So, the required solution of the equation is
$x = \left( {2, - 4,\left( { - 1 \pm \sqrt {71} } \right)} \right)$
So, this is the required answer.

Note: In such types of questions start with normal multiplication and then replace some portion of the equation with another variable and convert it into a quadratic equation, now factorize the equation if possible if not apply quadratic formula, so after simplification we will get the required answer.