Question

The zeros of a quadratic polynomial x^2 + 88 x + 125 are ?

Answer 1

Step-by-step explanation:

just apply the quadratic formula

\begin{lgathered}d = {b}^{2} - 4ac \\ and \: then \\ roots = \frac{ - b + - \sqrt{d} }{2a}\end{lgathered}

d=b

2

−4ac

andthen

roots=

2a

−b+−

d

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Answer 2

Step-by-step explanation:

x^2 + 88 x + 125

a=1

b=88

c=125

x=[-b±√(b²-4ac)]/2a=-88±√7244/2

so zeroes are -88+√7244/2 and -88-√7244/2