Answer
$14-i\sqrt 2$
Work Step by Step
Note: the standard form of a complex number is $a+bi$ where $i = \sqrt -1$ and $a$ and $b$ are real numbers.
Note: $i^{2} = \sqrt -1\sqrt -1 = -1$
$(\sqrt 2 + 3i)(\sqrt 2 - 4i)$
First, expand out the complex number:
$(\sqrt 2)^{2} - 4i\sqrt 2+3i\sqrt 2 - 12i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(\sqrt 2)^{2} - 4i\sqrt 2+3i\sqrt 2 - 12(-1)$
Simplify:
$(\sqrt 2)^{2} - 4i\sqrt 2+3i\sqrt 2 + 12$
$2 - 4i\sqrt 2+3i\sqrt 2 + 12$
Combine like terms, to express the complex number in standard form:
$14-i\sqrt 2$
