Question

The zeros of the quadratic polynomial x2+99x+127 are

Answer 1

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Answer 2

Ans  -----  ⇅

Step-by-step explanation:

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