Answer
$x=\frac{\sqrt {2}}{2}$ or $x=-2\sqrt {2}$
Work Step by Step
$\sqrt {2} x^2 + 3x - 2\sqrt{2} = 0$
Solve by using the quadratic formula:
$x=\frac{−b±\sqrt{b^2−4ac}}{2a}$
$a=\sqrt {2}$, $b=3$, $c=- 2\sqrt{2}$
$x=\frac{−3±\sqrt{(-3)^2−(4⋅\sqrt {2}⋅- 2\sqrt{2})}}{2⋅\sqrt {2}}$
$x=\frac{−3±\sqrt{9−(-16)}}{2\sqrt {2}}$
$x=\frac{−3±\sqrt{25}}{2\sqrt {2}}$
$x=\frac{−3±5}{2\sqrt {2}}$
$x=\frac{−3+5}{2\sqrt {2}}$ or $x=\frac{−3-5}{2\sqrt {2}}$
$x=\frac{2}{2\sqrt {2}}$ or $x=\frac{−8}{2\sqrt {2}}$
$x=\frac{\sqrt {2}}{2}$ or $x=-2\sqrt {2}$
