Inequalities question I've been having trouble solving these kind of inequalities ; 
$\sqrt { -7x+1 } - \sqrt {x+10}  \gt \  {7}$ 
Attempt at a Solution; We first find the official boundaries of the inequality; x is limited to the section:  $\ -10 \le x \le \frac{1}{7} $
Then, squaring both sides we get: 
${-7x+1} -2 \sqrt{(x+10)}\sqrt{(-7x+1)} +x+10 \gt 49  $
$-2 \sqrt{(x+10)}\sqrt{(-7x+1)}  \gt 38+6x $
squaring again eventually yields:
$ -28x^2-276x+40\gt 1444+456x+36x^2 $
$-64x^2 -732x -1404 \gt 0 $
binomial roots are as specified below. 
since the parabola is concave and in order this inequality is to be satisfied, 
x must assume the values:$\ -9 \lt x \lt  -2.4375 $ upon intersecting this with the range we get the same result and this is the (apparent) final solution. 
We then resume to the usual manipulation of the inequality; squaring both sides, factoring etc. We get two roots; $-9$,$ -2.4375$ . The parabola we get after these manipulations is concave, so we get the solution:  $\ -9 \lt x \lt  -2.4375 $ .  We then intersect this solution with the range for x, yielding the final solution  $\ -9 \lt x \lt  -2.4375 $. Upon testing it we find it does not fulfill the inequality we tried to solve, so we cancel it as a solution, leading to that there are no solutions to this inequality, as a final answer. Official answers state otherwise.
Any help? 
 A: From $\sqrt{-7x+1} - \sqrt{x+10}  \gt 7$, your assessment of the interval $x \in [-10, \frac{1}{7}]$is clearly right.
Let us make sure we have positive quantities on both sides first, so rewrite as
$\sqrt{-7x+1}  \gt 7 + \sqrt{x+10} $  
Squaring,  $- 7x + 1 > 49 + x + 10 + 14\sqrt{x+10}  $
or $-4x -29 > 7\sqrt{x+10} $
Now it is easier to see that the LHS is positive only when $x < -\frac{29}{4}$, so the allowable values have shrunk to $x \in [-10, -\frac{29}{4}]$.  In this interval, we have assurance of both sides being positive, so squaring again, we have
$16x^2 + 232x + 841 > 49(x+10) \implies 16x^2 + 183x + 351 > 0$  So we have
$$16\left(x+\dfrac{183}{32}\right)^2 > \dfrac{11025}{64} \implies x +\dfrac{183}{32} \not \in \left[- \frac{105}{32}, \frac{105}{32} \right]  $$
Thus the allowable values are further shrunk to $x \in [-10, -9)$.
A: First of all, before squaring the initial inequality, it seems appropriate to note that $\sqrt{-7x+1}-\sqrt{x+10}$ has to be greater that zero (i don't know, perhaps this is redundant, but i'd do that just to make sure...)
Then, after squaring both parts, i got $3x+19 + \sqrt{(-7x+1)(x+10)} < 0$, which gives us $\sqrt{(-7x+1)(x+10)} <-(3x+19)$, if i'm not mistaken. This yelds the condition $3x+19 < 0$, or $x < -\frac{19}{3}$. Then i squared everything, and got almost what you got - $x < -9$ or $x > -\frac{78}{32}$. So now it is enough to intersect all inequalities properly... Will the answer be correct?
And my final answer seems to be $[-10,-9)$ (after calculations)... Is it correct?
