Question

How to find the vertex of the,quadratic equation m^2-5m-14=0

Answer 1

$quadratic\ formula:ax^2+bx+c=0\\\\m^2-5m-14=0\\\\a=1;\ b=-5;\ c=-14\\\\the\ vertex:(p;\ q)\ where\ p=\frac{-b}{2a}\ and\ q=-\frac{b^2-4ac}{4a}\\\\p=\frac{-(-5)}{2\cdot1}=\frac{5}{2}=2.5\\\\q=-\frac{(-5)^2-4\cdot1\cdot(-14)}{4\cdot1}=-\frac{25+56}{4}=-\frac{81}{4}=-20.25\\\\Answer:The\ vetrex\ is\ (2.5;-20.25)$