To determine the zeros of the quadratic function [tex]\( f(x) = 2x^2 - 10x - 3 \)[/tex], we need to solve for the values of [tex]\( x \)[/tex] where [tex]\( f(x) = 0 \)[/tex].
We start by using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( b = -10 \)[/tex], and [tex]\( c = -3 \)[/tex].
First, we calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-10)^2 - 4(2)(-3) \][/tex]
[tex]\[ \Delta = 100 + 24 \][/tex]
[tex]\[ \Delta = 124 \][/tex]
Now that we have the discriminant, we can find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{10 \pm \sqrt{124}}{4} \][/tex]
Simplify [tex]\(\sqrt{124}\)[/tex]:
[tex]\[ \sqrt{124} = 2\sqrt{31} \][/tex]
Thus, the equation for the roots becomes:
[tex]\[ x = \frac{10 \pm 2\sqrt{31}}{4} \][/tex]
[tex]\[ x = \frac{10 \pm 2\sqrt{31}}{4} = \frac{10}{4} \pm \frac{2\sqrt{31}}{4} \][/tex]
[tex]\[ x = \frac{5}{2} \pm \frac{\sqrt{31}}{2} \][/tex]
So the zeros of the quadratic function [tex]\( f(x) = 2x^2 - 10x - 3 \)[/tex] are:
[tex]\[ x = \frac{5}{2} - \frac{\sqrt{31}}{2} \][/tex]
[tex]\[ x = \frac{5}{2} + \frac{\sqrt{31}}{2} \][/tex]
Therefore, the correct answer from the given options is:
[tex]\[ x=\frac{5}{2}-\frac{\sqrt{31}}{2} \text{ and } x=\frac{5}{2}+\frac{\sqrt{31}}{2} \][/tex]
