# How do you find the zeros of the function f(x)=(20x^2+11x−40)/(2x+5)?

How do you find the zeros of the function $f\left(x\right)=\frac{20{x}^{2}+11x-40}{2x+5}$?
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Simeon Hester
the denominator of f(x) cannot be zero as this would make f(x) undefined
the zeros are found by equating the numerator to zero
$⇒\text{solve}\phantom{\rule{1ex}{0ex}}20{x}^{2}+11x-40=0←\text{standard form}$
with a=20,b=11 and c=−40
$•xx=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$⇒x=\frac{-11±\sqrt{121+3200}}{40}$
$⇒x=\frac{-11±\sqrt{3321}}{40}=\frac{-11±9\sqrt{41}}{40}$
$⇒x=-\frac{11}{40}±\frac{9\sqrt{41}}{40}$
$⇒x=-\frac{11}{40}-\frac{9\sqrt{41}}{40}\phantom{\rule{1ex}{0ex}}\text{or}\phantom{\rule{1ex}{0ex}}x=-\frac{11}{40}+\frac{9\sqrt{41}}{40}$
$⇒\text{zeros are}\phantom{\rule{1ex}{0ex}}x\approx -1.72,x\approx 1.17\phantom{\rule{1ex}{0ex}}\text{to 2 dec. places}$