# What are the solutions of the quadratic equation 0 = 4(x − 3)^{2} − 16 ?

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

## Answer: The solution set of the following equation 0 = 4(x − 3)^{2} − 16 is 5 and 1.

Let's find the values of x.

**Explanation:**

Given: 0 = 4(x − 3)^{2} − 16

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

ax^{2} + bx + c = 0

To convert the equation into standard form,

⇒ 0 = 4(x − 3)^{2} − 16

⇒ 0 = 4 (x^{2} - 6x + 9 ) − 16

⇒ 0 = 4 x^{2} - 24x + 36 - 16

⇒ 4 x^{2} - 24x + 20 = 0

- the coefficient of x
^{2 }= 4 - the coefficient of x = - 24
- constant term = 20

Sum = - 24 , Product = 4 × 20 = 80

Thus, by splitting the middle term,

⇒ 4 x^{2} - 4 x - 20 x + 20 = 0

⇒ 4 x ( x - 1 ) - 20 ( x - 1 ) = 0

⇒ ( 4 x - 20 ) ( x - 1 ) = 0

⇒ 4 x - 20 = 0 or x - 1 = 0

⇒ x = 5 or 1