Question

find the zeroes of the quadratic polynomial p(X) = 6x^2 - 3-7x??​

Answer 1

Answer:

-1/3, 3/2

Step-by-step explanation:

Answer 2

Answer:

$6 {x}^{2} - 3 - 7x \\ 6 {x}^{2} - 7x - 3 \\ 6 {x}^{2} + 2x - 9x - 3 \\ 2x(3x + 1) \: - 3(3x + 1) \\ (2x - 3) + (3x + 1) = 0 \\ 2x - 3 = 0 \: \: 3x + 1 = 0 \\ 2x = 3 \: and \: 3x = - 1 \\ x = \frac{3}{2} \: \: \: x = \frac{ - 1}{3}$