Question 353070
The equation's already in vertex form, {{{y=a(x-h)^2+k}}} so then the vertex,({{{h}}},{{{k}}})=({{{-2}}},{{{-3}}})
To find the zeros, set {{{y=0}}} and solve.
{{{(x+2)^2-3=0}}}
{{{(x+2)^2=3}}}
{{{x+2=0 +- sqrt(3)}}}
{{{highlight(x=-2 +- sqrt(3))}}}
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{{{drawing(300,300,-8,6,-7,7,blue(line(-2,10,-2,-10)),circle(-2+sqrt(3),0,0.35),circle(-2-sqrt(3),0,0.35),circle(-2,-3,0.35),graph(300,300,-8,6,-7,7,(x+2)^2-3))}}}