The problem asks us to solve the system of equations: $x^2 + y^2 = 3$ $x^2 - y = 15$

1. Problem Description

The problem asks us to solve the system of equations:

$x^2 + y^2 = 3$

$x^2 - y = 15$

2. Solution Steps

We have two equations:

$x^2 + y^2 = 3$ (1)

$x^2 - y = 15$ (2)

From equation (2), we can express $x^2$ in terms of $y$:

$x^2 = y + 15$ (3)

Substitute equation (3) into equation (1):

$(y + 15) + y^2 = 3$

$y^2 + y + 15 = 3$

$y^2 + y + 12 = 0$

We can use the quadratic formula to solve for $y$:

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

where $a = 1$, $b = 1$, and $c = 12$.

$y = \frac{-1 \pm \sqrt{1^2 - 4(1)(12)}}{2(1)}$

$y = \frac{-1 \pm \sqrt{1 - 48}}{2}$

$y = \frac{-1 \pm \sqrt{-47}}{2}$

$y = \frac{-1 \pm i\sqrt{47}}{2}$

Let $y_1 = \frac{-1 + i\sqrt{47}}{2}$ and $y_2 = \frac{-1 - i\sqrt{47}}{2}$.

Substitute $y_1$ into equation (3):

$x^2 = y_1 + 15 = \frac{-1 + i\sqrt{47}}{2} + 15 = \frac{-1 + i\sqrt{47} + 30}{2} = \frac{29 + i\sqrt{47}}{2}$

So, $x = \pm \sqrt{\frac{29 + i\sqrt{47}}{2}}$

Substitute $y_2$ into equation (3):

$x^2 = y_2 + 15 = \frac{-1 - i\sqrt{47}}{2} + 15 = \frac{-1 - i\sqrt{47} + 30}{2} = \frac{29 - i\sqrt{47}}{2}$

So, $x = \pm \sqrt{\frac{29 - i\sqrt{47}}{2}}$

3. Final Answer

The solutions are:

$y = \frac{-1 \pm i\sqrt{47}}{2}$

$x = \pm \sqrt{\frac{29 \pm i\sqrt{47}}{2}}$

The solutions are complex.

$x = \pm \sqrt{\frac{29 + i\sqrt{47}}{2}}, y = \frac{-1 + i\sqrt{47}}{2}$

$x = \pm \sqrt{\frac{29 - i\sqrt{47}}{2}}, y = \frac{-1 - i\sqrt{47}}{2}$