How to solve $(\sqrt{x}+7)(\sqrt{x}-1)=\frac{105}{4}$? I'm having difficulties solving $(\sqrt{x}+7)(\sqrt{x}-1)=\frac{105}{4}$. I guess I'm missing a crucial point. The solution is $x=\frac{49}{4}$. What are the steps?
 A: Let $t = \sqrt x$.
Then we can convert the question to
$(t + 7)(t - 1) = \frac{105}{4}$
Simplify it to get the following quadratic in $t$
$4t^2 + 24t - 133 = 0$
There are many ways to solve a quadratic.
By any method, you will get the following solutions for $t$
$t_1 = \frac{7}{2}$
$t_2 = \frac{-19}{2}$
From here, we can solve for $x $ to get the desired answer. Note that we want $t$ to be positive therefore we do not consider $t_2$.
A: As the user Dark Magician said, we can use the substitution $t=\sqrt{x}$. We get the following from using the quadratic formula:
\begin{align}
(t+7)(t-1)&=\frac{105}{4}\\
t^2-t+7t-7&=\frac{105}{4}\\
t^2+6t-7-\frac{105}{4}&=0\\
t^2+6t-\frac{133}{4}&=0\\
4t^2+24t-133&=0\\
t&=\frac{-24\pm\sqrt{24^2-4\times4\times(-133)}}{2\times4}\\
t&=\frac{-24\pm\sqrt{576+2128}}{8}\\
t&=\frac{-24\pm\sqrt{2704}}{8}\\
t&=-3\pm\frac{13}{2}\\
t&=-\frac{19}{2}\ or\ \frac{7}{2}
\end{align}
We know that $t$ is nonnegative, as it is an arithmetic square root, so
$$t=\frac{7}{2}$$
$$x=t^2=\frac{49}{4}$$
A: The question setter has the intention of requiring a "$t = \sqrt x + 3$" substitution such that the original becomes $(t + 4) (t - 4) = 105/14$.
Then, $t^2 - 4^2 = 105/4$
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