Solve for ‘x’ From an Algebra Math Equation

4 \sqrt{\dfrac{3-x}{3+x}} -  \sqrt{\dfrac{3+x}{3-x}} = \sqrt2

x = ??

\begin{aligned}
4 \sqrt{\dfrac{3-x}{3+x}} -  \sqrt{\dfrac{3+x}{3-x}} &= \sqrt2\\
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4 \sqrt{\dfrac{3-x}{3+x}} -  \sqrt{\dfrac{3+x}{3-x}}  &={\dfrac{4\sqrt{3-x}}{\sqrt{3+x}}} -  {\dfrac{\sqrt{3+x}}{\sqrt{3-x}}} \\
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&={\dfrac{4({3-x})-({3+x})}{\sqrt{9-x^2}}} \\
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&={\dfrac{12-4x-3-x}{\sqrt{9-x^2}}} \\
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&={\dfrac{9-5x}{\sqrt{9-x^2}}} \\
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\Rightarrow {\dfrac{9-5x}{\sqrt{9-x^2}}} &= \sqrt{2}\\
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\Rightarrow {9-5x}&= \sqrt{18-2x^2}\\
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\end{aligned}

\begin{aligned}
({9-5x})^2&={18-2x^2}\\
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\Rightarrow 81-90x+25x^2&={18-2x^2}\\
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\Rightarrow 27x^2+63-90x &=0\\
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\Rightarrow 27x^2-90x+63 &=0\\
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\Rightarrow 3x^2-10x+7 &=0\\
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\end{aligned}

\begin{aligned}
x &=\frac{10 \pm \sqrt{10^2-4 \times 3 \times 7}}{2 \times 3}\\
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&=\frac{10 \pm \sqrt{100-84}}{6}\\
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&=\frac{10 \pm 4}{6}\\
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\Rightarrow x&=\frac{7}{3}, 1\\
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\end{aligned}