Answer
The solutions are $x=-2\sqrt{2}$ and $x=\dfrac{\sqrt{2}}{2}$
Work Step by Step
$\sqrt{2}x^{2}+3x-2\sqrt{2}=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=\sqrt{2}$, $b=3$ and $c=-2\sqrt{2}$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-3\pm\sqrt{3^{2}-4(\sqrt{2})(-2\sqrt{2})}}{2\sqrt{2}}=\dfrac{-3\pm\sqrt{9+16}}{2\sqrt{2}}=...$
$...=\dfrac{-3\pm\sqrt{25}}{2\sqrt{2}}=\dfrac{-3\pm5}{2\sqrt{2}}$
The two solutions are:
$x=\dfrac{-3+5}{2\sqrt{2}}=\dfrac{2}{2\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}$
$x=\dfrac{-3-5}{2\sqrt{2}}=\dfrac{-8}{2\sqrt{2}}=-\dfrac{4}{\sqrt{2}}=-\dfrac{4\sqrt{2}}{2}=-2\sqrt{2}$
