Solve the equation $(x-3)(x+9)(x+5)(x-7)=385$ I have one question please...I solved it in this way..so I am not sure..that is it right or not? and If I solved it in the wrong way..so would like to know about the correct way and method to solve it..so please :
Solve the equation (x-3) (x+9) (x+5) (x-7) = 385

and here is I solved it in this way : 
Solve the equation:
(x-3) (x+9) (x+5) (x-7) = 385
x ( (1-3) (1+9) (1+5) (1-7)) = 385
x ( (-2) (10) (6) (-6) ) = 385
x ( (-20) (6) (-6) ) = 385
x ( (-120) (-6) ) = 385
x ( 720 ) = 385
720x = 385
x = 385/720
x = 0.5347222

waiting for your replies...
 A: Observe that  $\displaystyle -3+5=-7+9$
So, $$(x-3)(x+5)=x^2+2x-15, (x+9)(x-7)=x^2+2x-63$$
Set $\displaystyle x^2+2x=u$ to find $\displaystyle (u-15)(u-63)=385$
$\displaystyle\implies u^2-78u+560=0$
Using Quadratic Equations  Formula, $\displaystyle u=8,60$
or using Middle Term Factor, $\displaystyle u^2-78u+560=u^2-(70+8)u+70\cdot8=(u-70)(u-8)$
So, we have $\displaystyle x^2+2x=8$ or $\displaystyle x^2+2x=70$
Can you take it from here?
A: Some easy ideas to attack your problem with integer solutions: 
$$(x-3)(x+9)(x+5)(x-7)=385=5\cdot7\cdot11\implies$$
one of the factors in the LHS must be $\;\pm1\;$ , and after a little action we can see this happens for $\;x=2\;$ since the signs fit nicely (two positive, two negative):
$$(2-3)(2+9)(2+5)(2-7)=(-1)\cdot11\cdot7\cdot(-5)=385$$
Try now also $\;x=-4\;$ ...
A: Let $y = x + 1$ (I got this by taking the average of the four). Then the problem becomes:
$$(y-4)(y+8)(y+4)(y-8) = 385$$
which can be simplified nicely (by multiplying out similar terms) into:
$$\begin{align}(y^2 - 16)(y^2 - 64) &= 385 \\&= 7\cdot55\\&=-7\cdot-55\end{align}$$
The difference between $7$ and $55$ is $48$. But this is exactly the difference between $y^2 - 16$ and $y^2 - 64$. Hence we equate them according to their magnitudes ($y^2 - 16 = 55$) to deduce that:
$$\begin{align}&y^2 - 16 = 55 \\\implies&y^2 = 71\\\implies &(x+1)^2 = 71 \\ \implies &x = -1\pm\sqrt{71}\end{align}$$
Now, we deal with $-7$ and $-55$ in a similar manner:
$$\begin{align}&y^2 - 16 = -7 \\\implies&y^2 = 9\\\implies &(x+1)^2 = 9 \\ \implies &x = -1\pm3\end{align}$$
We conclude with with our four solutions for $x$:
$$x = -1\pm\sqrt{71}, -1\pm3$$
The best part: No quadratic formula used!
