```Question 320057
The equation is already in vertex form, {{{y=a(x-h)^2+k}}}, where (h,k) is the vertex.
Comparing
{{{f(x)=-3(x+2)^2}}}
{{{f(x)=-3(x+2)^2+0}}}
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(h,k)=(-2,0).
The axis of symmetry includes the vertex.
{{{x=-2}}}
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{{{drawing(300,300,-5,5,-5,5,grid(1),circle(-2,0,.3),blue(line(-2,-10,-2,10)),graph(300,300,-5,5,-5,5,-3(x+2)^2))}}}
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{{{4x^2=-9x-2}}}
{{{4x^2+9x+2=0}}}
{{{4(x^2+(9/4)x)+2=0}}}
{{{4(x^2+(9/4)x+81/64)+2-81/16=0}}}
{{{4(x+9/8)^2-49/16=0}}}
{{{4(x+9/8)^2=49/16}}}
{{{(x+9/8)^2=49/64}}}
{{{x+9/8=0 +- 7/8}}}
{{{x=-9/8 +- 7/8}}}
{{{highlight(x=-2)}}} and {{{highlight(x=-1/4)}}}
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{{{x^2 + (13/2)x =13}}}
{{{x^2 + (13/2)x +(13/4)^2=13+(13/4)^2}}}
{{{(x+(13/4))^2=208/16+169/16}}}
{{{(x+(13/4))^2=377/16}}}
{{{x+13/4=0 +- sqrt(377)/4}}}
{{{highlight(x=-13/4 +- sqrt(377)/4)}}}
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For a rectangle,
{{{A=L*W=15}}}
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{{{W=L-2}}}
Substitute,
{{{L*(L-2)=15}}}
{{{L^2-2L=15}}}
{{{L^2-2L-15=0}}}
{{{(L-5)(L+3)=0}}}
Only the positive solution makes sense in this problem,
{{{L-5=0}}}
{{{highlight(L=5)}}}
Then,
{{{W=L-2}}}
{{{W=5-2}}}
{{{highlight(W=3)}}}
```