I am quite sure EP will be correct but I will take a look:
Let b and c be constants such that the quadratic \(-2x^2 +bx +c\)
has roots \(3+\sqrt{5}\) and \(3-\sqrt{5}\).
Find the vertex of the graph of the equation
\(y=-2x^2 + bx + c\)
The roots of a quadratic \( ax^2+bx+c\) is the answers to \( ax^2+bx+c=0\)
And the answers to this are given by the quadratic equation
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=3\pm\sqrt5 \)
This means that \(x=\frac{-b}{2a}\) = 3 must be exactly halfway between the two roots.!!
So the axis of symmetry is x=3 and the vertex lies on this line so the x value of the vertex is 3
The y value will be \(y=-2*3^2+bx+c = -18+3b+c \)
This is fine but you need to find the vleu of b and c 
From the equation above I can see that
\(\frac{-b}{2a}=3 \qquad and \qquad \pm\frac{\sqrt{b^2-4ac}}{2a}=\pm\sqrt5\\ a=-2\qquad so\\ \frac{-b}{-4}=3 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{-4}=\pm\sqrt5\\ b=12 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{4}=\pm\sqrt5\\ b=12 \qquad and \qquad \frac{b^2+8c}{16}=5\\ b=12 \qquad and \qquad b^2+8c=80\\ b=12 \qquad and \qquad 144+8c=80\\ b=12 \qquad and \qquad 18+c=10\\ b=12 \qquad and \qquad c=-8\\\)
so
\(y= -18+3b+c\\ y= -18+3*12+-8\\ y=-18+36-8\\ y=10 \)
So the vertex is (3,10)
Which is exactly the same answer has given you. He has done it a number of different ways but stlill all the answers are the same!!
Thanks EP :)
