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find the zeros of the function f(x)=2x^2-7x-30

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[tex]2x^2-7x-30=0 \\ \\ a=2 \\ b=-7 \\ c=-30 \\ b^2-4ac=(-7)^2-4 \times 2 \times (-30)=49+240=289 \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-7) \pm \sqrt{289}}{2 \times 2}=\frac{7 \pm 17}{4} \\ x=\frac{7-17}{4} \ \lor \ x=\frac{7+17}{4} \\ x=\frac{-10}{4} \ \lor \ x=\frac{24}{4} \\ x=-\frac{5}{2} \ \lor \ x=6 \\ \boxed{x=-2.5 \hbox{ or } x=6}[/tex]
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