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=(-b)/(2a)\ and\ q=-(b^2-4ac)/(4a)\\\\p=(-(-5))/(2\cdot1)=(5)/(2)=2.5\\\\q=-((-5)^2-4\cdot1\cdot(-14))/(4\cdot1)=-(25+56)/(4)=-(81)/(4)=-20.25\\\\Answer:The\ vetrex\ is\ (2.5;-20.25)`