Question 925625
Substituting,
{{{x^2+4x=12}}}
{{{x^2+4x-12=0}}}
{{{(x+6)(x-2)=0}}}
Two solutions:
{{{x+6=0}}}
{{{x=-6}}}
{{{y^2=-24}}}
No real y solutions.
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{{{x-2=0}}}
{{{x=2}}}
Then,
{{{y^2=8}}}
{{{y=0 +- sqrt(8)}}}
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({{{2}}},{{{sqrt(8)}}})
({{{2}}},{{{-sqrt(8)}}})
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{{{drawing(300,300,-4,6,-4,6,grid(1),circle(2,sqrt(8),0.2),circle(2,-sqrt(8),0.2),graph(300,300,-4,6,-4,6,sqrt(12-x^2),-sqrt(12-x^2),sqrt(4x),-sqrt(4x)))}}}