help in solving a quadratic equation

[Oldguy]

Hi, I have not been in school for more than 35 years, so I am having a hard time getting my brain to remember even on how to start. If somebody could help in getting me started with ths equation, or even show me the answer and how you got it. I have been trying to find a simple explanation online but I can't seem to find one in laymans terms. I am trying to answer a sample question that is on an upgrade in my trade book. It's the only one I can't answer correctly. thanks in advance for any help and hopefully I have put this in the right category, Merry Christmas.


Find the vertex of the quadratic equation: x²-132x= -4358. Answer in (x,y) format

 

[galactus]

If you have a quadratic in the form \(\displaystyle ax^{2}+bx+c=0\), then the vertex coordinates are given by

\(\displaystyle x=\frac{-b}{2a}, \;\ y=c-\frac{b^{2}}{4a}\)

Rewrite yours so it is \(\displaystyle x^{2}-135x+4358=0\)

Now, simply apply the formulas. \(\displaystyle a=1, \;\ b=-135, \;\ c=4358\)

 

[Oldguy]

The answer ?

Hi , thanks to Galactus and JeffM om their replys, you got me on the right path, galactus, in your reply you put -135X, you did mean -132X right? I think I have the answer, so I am asking if it is right, if not please answer it and show me the workings.

Find the vertex of the quadratic equation: x²-132x= -4358. Answer in (x,y) format

Answer ? X= 66 Y=2

thanks