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