# What is the solution set of the following equation 6x^{2 }= 13x + 5 ?

An equation in the form of ax^{2} + bx + c = 0 is called as a quadratic equation.

## Answer: The solution set of the following equation 6x^{2 }= 13x + 5 is - 1/3 or 5/2.

Let's find the values of x.

**Explanation:**

We know that the standard form of a quadratic equation is given by,

ax^{2} + bx + c = 0

Thus, the quadratic equation 6x^{2 }= 13x + 5 can be written as 6x^{2 } - 13x - 5 = 0

In 6x^{2 } - 13x - 5 = 0,

- the coefficient of x
^{2 }= 6 - the coefficient of x = - 13
- constant term = - 5.

Sum = - 13, Product = 6 × (- 5) = - 30

Thus, by splitting the middle term,

⇒ 6 x^{2 } + 2 x - 15 x - 5 = 0

⇒ 2x (3x + 1) - 5 (3x + 1) = 0

⇒ (3x + 1) ( 2x - 5 ) = 0

⇒ x = - 1/3 or 5/2

Thus, the value of x that satisfies the equation 6x^{2 }= 13x + 5 is - 1/3 or 5/2.