Question 1052785
Covert the equation to vertex form,
{{{y=ax^2+bx+c}}}
{{{y=a(x^2+(b/a)x)+c}}}
{{{y=a(x^2+(b/a)x+(b/(2a))^2)+c-a(b/(2a))^2}}}
{{{y=a(x+(b/(2a)))^2+(c-b^2/(4a))}}}
So then the vertex(turning point) is 
({{{-b/(2a)}}},{{{c-b^2/(4a)}}})
So then for the example,
{{{a=-2}}}
{{{b=-2}}}
{{{c=3}}}
The turning point would be,
({{{-(-2/(2(-2)))}}},{{{3-(-2)^2/(4(-2))}}})
({{{-1/2}}},{{{7/2}}})