Question

Solve the following equation using the general formula:
$x^2-3\sqrt{6}x+12=0$

Answer 1

$x^2-3\sqrt{6}x+12=0$

$x=-\frac{b\pm \sqrt{b^2-4ac}}{2a}$

$x=-\frac{(-3\sqrt{6})\pm \sqrt{(3\sqrt{6}{)}^2-4\times 1\times 12}}{2\times 1}$

$=\frac{3\sqrt{6}\pm \sqrt{54-48}}{2}$

$=\frac{3\sqrt{6}\pm 2}{2}$

$x=\frac{3\sqrt{6}+2}{2}$, $\frac{3\sqrt{6}-2}{2}$.