Answer
$$x=\dfrac{y\pm \sqrt{8-11y^2}}{4}$$
$$y=\dfrac{x\pm \sqrt{6-11x^2}}{3}$$
Work Step by Step
To solve for $x$ in terms of $y$, first rewrite in standard quadratic equation form: $ax^2 + bx+c=0$
$4x^2-2xy+3y^2-2$
where $a=4$, $b=-2y$, and $c=(3y^2-2)$
now, apply the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{(-)(-2y)\pm \sqrt{(-2y)^2-4(4)(3y^2-2)}}{2(4)}$
$x=\dfrac{2y\pm \sqrt{4y^2-16(3y^2-2)}}{8}$
$x=\dfrac{2y\pm \sqrt{4y^2-48y^2+32)}}{8}$
$x=\dfrac{2y\pm \sqrt{32-44y^2)}}{8}$
$x=\dfrac{2y\pm \sqrt{(4)(8-11y^2)}}{8}$
$x=\dfrac{2y\pm 2\sqrt{8-11y^2}}{8}$
$$x=\dfrac{y\pm \sqrt{8-11y^2}}{4}$$
To solve for $y$ in terms of $x$, first rewrite in standard quadratic equation form: $ay^2 + by+c=0$
$3y^2-2xy+4x^2-2=0$
where $a=3$, $b=-2x$, and $c=(4x^2-2)$
now, apply the quadratic formula $y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$y=\dfrac{(-)(-2x)\pm \sqrt{(-2x)^2-4(3)(4x^2-2)}}{2(3)}$
$y=\dfrac{2x\pm \sqrt{4x^2-12(4x^2-2)}}{6}$
$y=\dfrac{2x\pm \sqrt{4x^2-48x^2+24)}}{6}$
$y=\dfrac{2x\pm \sqrt{24-44x^2)}}{6}$
$y=\dfrac{2x\pm \sqrt{4(6-11x^2)}}{6}$
$y=\dfrac{2x\pm 2\sqrt{6-11x^2}}{6}$
$$y=\dfrac{x\pm \sqrt{6-11x^2}}{3}$$
