Zeroes of polynomial p(x)=2x^2-9-3x are :​

x = - , x = 3

Step-by-step explanation:

To find the zeros let p(x) = 0 , that is

2x² - 3x - 9 = 0

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 2 × - 9 = - 18 and sum = - 3

The factors are - 6 and + 3

Use these factors to split the x- term

2x² - 6x + 3x - 9 = 0 ( factor the first/second and third/fourth terms)

2x(x - 3) + 3(x - 3) = 0 ← factor out (x - 3) from each term

(x - 3)(2x + 3) = 0

Equate each factor to zero and solve for x

2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = -

x - 3 = 0 ⇒ x = 3

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15