[tex]\bf \begin{cases}
-2x^2+y=-5\\
\boxed{y}=-3x^2+5\\
----------\\
-2x^2+\left( \boxed{-3x^2+5} \right)=-5
\end{cases}
\\\\\\
-5x^2+5=-5\implies -5(x^2-1)=-5\implies x^2-1=\cfrac{-5}{-5}
\\\\\\
x^2-1=1\implies x^2=2\implies x=\pm\sqrt{2}[/tex]
and since we know what "x" is, then let's substitute it on say the 1st equation
[tex]\bf -2x^2+y=-5\implies -2\left( \pm \sqrt{2} \right)^2+y=-5\implies -4+y=-5
\\\\\\
y=-1\\\\
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\left( \sqrt{2}~,~-1 \right)\qquad \left( -\sqrt{2}~,~-1 \right)[/tex]
