Question

what are the zeros of the quadratic function ( f(x) = 2x^2 - 10x - 3 )?

Explanation:

Step1: Identify quadratic coefficients

For $f(x)=2x^2-10x-3$, $a=2$, $b=-10$, $c=-3$

Step2: Apply quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Step3: Calculate discriminant

$\Delta = (-10)^2 - 4(2)(-3) = 100 + 24 = 124$

Step4: Simplify discriminant root

$\sqrt{124}=\sqrt{4\times31}=2\sqrt{31}$

Step5: Substitute into formula

$x=\frac{10\pm2\sqrt{31}}{4}=\frac{5\pm\sqrt{31}}{2}$
Rewrite as: $x=\frac{5}{2}+\frac{\sqrt{31}}{2}$ and $x=\frac{5}{2}-\frac{\sqrt{31}}{2}$

Answer:

$x = \frac{5}{2} - \frac{\sqrt{31}}{2}$ and $x = \frac{5}{2} + \frac{\sqrt{31}}{2}$