Area of ABC : ABAC/2
the maximum of the parabola is reached at x=-4/(2(-1))=2 hence A is at (2,0) and B is at (2,(-2)^2+42+C)=(2,12+C)
C is the second root (x-intersect), which we can find :
determinant1 : D=16-4(-1)*C=4(4+C) thus the second root is at x=\(\frac{-4-\sqrt{4(4+C)}}{-2}=2+\sqrt{4+C}\)
Hence the area of the triangle is \(AB*AC/2=(4+C)*(2+\sqrt{4+C}-2)/2=(4+C)\sqrt{4+C}/2=32\) hence \((4+C)\sqrt{4+C}=64\) .
We remark that \(64=16*4=16*\sqrt{16}\)
Hence 4+C=16 thus C=12
