I am quite sure EP will be correct but I will take a look:
Let b and c be constants such that the quadratic −2x2+bx+c
has roots 3+√5 and 3−√5.
Find the vertex of the graph of the equation
y=−2x2+bx+c
The roots of a quadratic ax2+bx+c is the answers to ax2+bx+c=0
And the answers to this are given by the quadratic equation
x=−b±√b2−4ac2ax=−b2a±√b2−4ac2a=3±√5
This means that x=−b2a = 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∗32+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
−b2a=3and±√b2−4ac2a=±√5a=−2so−b−4=3and±√b2+8c−4=±√5b=12and±√b2+8c4=±√5b=12andb2+8c16=5b=12andb2+8c=80b=12and144+8c=80b=12and18+c=10b=12andc=−8
so
y=−18+3b+cy=−18+3∗12+−8y=−18+36−8y=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 :)