 # What are the Zeros of the Quadratic Function f(x) = 2x^2 – 10x – 3?

In the previous article, we have learned the ‘4 Best methods to find the Zeros of a Quadratic Function‘.

Now we understand this concept with the following example.

## What are the zeros of the quadratic function f(x) = 2x^2 – 10x – 3?

Answer: $x=\frac{5+\sqrt[]{31}}{2},\frac{5-\sqrt[]{31}}{2}$

### Detailed Solution:

The given quadratic function is $f(x) = 2x^2 – 10x – 3$.

The zeros of the quadratic function $f(x) = 2x^2 – 10x – 3$ are the values of x for which the value of the quadratic function $f(x) = 2x^2 – 10x – 3$ is equal to zero.

Equating $f(x) = 2x^2 – 10x – 3$ with $0$, we get the following equation,

$2x^2 – 10x – 3 = 0 …..\left ( 1 \right )$.

By the Formula of the Quadratic Equation, we know if $ax^2 + bx + c = 0 …..\left ( 2 \right )$, then the zeros are

$x= \frac{-b\pm \sqrt{b^{2}-4ac} }{2a}…..\left ( 3 \right )$

Comparing equations (1) and (2) we get,

$a = 2, b = - 10, c = - 3$.

Putting these values in equation (3) we get,

$x = \frac{ - \left ( - 10 \right ) \pm \sqrt{\left ( - 10 \right )^{2} - 4 \times 2 \times \left ( - 3 \right )}}{2 \times 2}$

or, $x = \frac{ 10 \pm \sqrt{ 100 + 24 }}{ 4 }$

or, $x = \frac{ 10 \pm \sqrt{ 124 }}{ 4 }$

or, $x = \frac{ 10 \pm 2 \sqrt{ 31 }}{ 4 }$

or, $x = \frac{ 5 \pm \sqrt{ 31 }}{ 2 }$

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

Therefore the zeros of the quadratic function $f(x) = 2x^2 – 10x – 3$ are $x = \frac{ 5 + \sqrt{ 31 }}{ 2 }$ and $x = \frac{ 5 - \sqrt{ 31 }}{ 2 }$.