Question

The zeroes of the quadratic polynomial x2+99x+127 are​

Answer 1

Answer:

Hey!!

x²+99x+127=0(let, if 'x' is a root of p(x))

as, x=[–b±√(b²–4ac)]/2a

=>x=[–99±√{(99)²–4(1)(127)}]/2(1)

=>x=[–99±√{9801–508}]/2

=>x=[–99±√9239]/2

x=[–99±96.4]/2

Therefore,

x=[–99–96.4]/2 or x=[–99+96.4]/2

x=–97.7 OR x=–1.3

Answer 2

Answer:

Step-by-step explanation: