Answer
The solutions are $(-3+\sqrt{15},6-2\sqrt{15})$ and $(-3-\sqrt{15},6+2\sqrt{15})$.
Work Step by Step
The given system is
$y+2x=0$ ...... (1)
$y=x^2+4x-6$ ...... (2)
Subtract equation (1) from equation (2).
$\Rightarrow y-(y+2x)=x^2+4x-6$
$\Rightarrow y-y-2x=x^2+4x-6$
Simplify.
$\Rightarrow -2x=x^2+4x-6$
Add $2x$ to each side.
$\Rightarrow -2x+2x=x^2+4x-6+2x$
Simplify.
$\Rightarrow 0=x^2+6x-6$
Add $15$ to each side.
$\Rightarrow 0+15=x^2+6x-6+15$
Simplify.
$\Rightarrow 15=x^2+6x+9$
Write the right terms as square of the polynomial.
$\Rightarrow 15=(x+3)^2$
Take square root on each side.
$\Rightarrow \pm\sqrt{15}=x+3$
Subtract $3$ from each side.
$\Rightarrow \pm\sqrt{15}-3=x+3-3$
Simplify.
$\Rightarrow -3\pm\sqrt{15}=x$
Substitute $-3+\sqrt{15}$ for $x$ in equation (1).
$\Rightarrow y+2(-3+\sqrt{15})=0$
Simplify.
$\Rightarrow y-6+2\sqrt{15}=0$
Solve for $y$.
$\Rightarrow y=6-2\sqrt{15}$
Substitute $-3-\sqrt{15}$ for $x$ in equation (1).
$\Rightarrow y+2(-3-\sqrt{15})=0$
Simplify.
$\Rightarrow y-6-2\sqrt{15}=0$
Solve for $y$.
$\Rightarrow y=6+2\sqrt{15}$
Hence, the solutions are $(-3+\sqrt{15},6-2\sqrt{15})$ and $(-3-\sqrt{15},6+2\sqrt{15})$.
