What are the zeros of the quadratic function \( f(x)=2 x^{2}-10 x-3 \) ? \( x=-\frac{5}{2}-\frac{\sqrt{31}}{2} \) and \( x=-\frac{5}{2}+\frac{\sqrt{31}}{2} \) \( x=-\frac{5}{2}-\sqrt{\frac{37}{8}} \) and \( x=-\frac{5}{2}+\sqrt{\frac{37}{8}} \) \( x=\frac{5}{2}-\frac{\sqrt{31}}{2} \) and \( x=\frac{5}{2}+\frac{\sqrt{31}}{2} \) \( x=\frac{5}{2}-\sqrt{\frac{37}{8}} \) and \( x=\frac{5}{2}+\sqrt{\frac{37}{8}} \)

To find the zeros of the quadratic function \( f(x) = 2x^2 - 10x - 3 \), you can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = -10 \), and \( c = -3 \). Plugging in the values gives you \( b^2 - 4ac = 100 + 24 = 124 \). Thus, the zeros are \( x = \frac{10 \pm \sqrt{124}}{4} = \frac{5 \pm \sqrt{31}}{2} \)! Looking at the options, the correct zeros are \( x = \frac{5}{2} - \frac{\sqrt{31}}{2} \) and \( x = \frac{5}{2} + \frac{\sqrt{31}}{2} \). Ever wondered how quadratic equations come into play in the real world? They're all around us! From projectile motion in sports to calculating the optimal price for goods in business, quadratics help model scenarios where things grow or shrink in a parabolic shape. Understanding their zeros can predict key outcomes, making it essential for fields like engineering, finance, and even rocket science! When solving quadratics, people often forget to carefully compute the discriminant \( b^2 - 4ac \). A common mistake is misapplying the square root in the quadratic formula—taking only the positive root without considering both options! Always double-check your calculations and remember that both positive and negative roots are essential for finding all possible solutions!