Answer
$$x=\dfrac{-2y\pm\sqrt{10y^2+4}}{2}$$
$$y=\dfrac{2x\pm\sqrt{10x^2-6}}{3}$$
Work Step by Step
To solve for $x$ in terms of $y$, first rewrite in standard quadratic equation form: $ax^2 + bx +c=0$
$2x^2+4xy-3y^2-2=0$
where $a=2$, $b=4y$, and $c=(-3y^2-2)$
now, apply the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\dfrac{(-)(4y)\pm\sqrt{(4y)^2-4(2)(-3y^2-2)}}{2(2)}$
$x=\dfrac{-4y\pm\sqrt{16y^2-8(-3y^2-2)}}{4}$
$x=\dfrac{-4y\pm\sqrt{16y^2+24y^2+16}}{4}$
$x=\dfrac{-4y\pm\sqrt{40y^2+16}}{4}$
$x=\dfrac{-4y\pm\sqrt{(4)(10y^2+4)}}{4}$
$x=\dfrac{-4y\pm2\sqrt{10y^2+4}}{4}$
$$x=\dfrac{-2y\pm\sqrt{10y^2+4}}{2}$$
To solve for $y$ in terms of $x$, first rewrite in standard quadratic equation form: $ay^2 + by +c=0$
$-3y^2+4xy+2x^2-2=0$
where $a=-3$, $b=4x$, and $c=(2x^2-2)$
now, apply the quadratic formula: $y=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
$y=\dfrac{(-)(4x)\pm\sqrt{(4x)^2-4(-3)(2x^2-2)}}{2(-3)}$
$y=\dfrac{-4x\pm\sqrt{16x^2+12(2x^2-2)}}{-6}$
$y=\dfrac{-4x\pm\sqrt{16x^2+24x^2-24}}{-6}$
$y=\dfrac{-4x\pm\sqrt{40x^2-24}}{-6}$
$y=\dfrac{-4x\pm\sqrt{(4)(10x^2-6)}}{-6}$
$y=\dfrac{-4x\pm2\sqrt{10x^2-6}}{-6}$
$y=\dfrac{-2x\pm\sqrt{10x^2-6}}{-3}$
$$y=\dfrac{2x\pm\sqrt{10x^2-6}}{3}$$
