Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

6x² - 3 - 7x

Let,

⇒ 6x² - 3 - 7x = 0

⇒ 6x² - 7x - 3 = 0

⇒ 6x² - 9x + 2x - 3 = 0

⇒ 3x(2x - 3) + 1(2x - 3) = 0

⇒ (3x + 1)(2x - 3) = 0

⇒ 3x + 1 = 0 or 2x - 3 = 0

⇒ 3x = -1 or 2x = 3

⇒ x = $-\dfrac{1}{3}$ or x = $\dfrac{3}{2}$

Sum of zeroes = $-\dfrac{1}{3} + \dfrac{3}{2} = \dfrac{-2 + 9}{6}= \dfrac{7}{6} = \dfrac{-(-7)}{6} = \dfrac{-\text{(Coefficient of x)}}{\text{(Coefficient of x}^2)}$

Product of zeroes = $-\dfrac{1}{3} \times \dfrac{3}{2} = -\dfrac{1}{2} = \dfrac{-3}{6} = \dfrac{\text{Constant term}}{\text{Coefficient of x}^2}$

Hence, for zero of the polynomial 6x² - 3 - 7x, x = $-\dfrac{1}{3}, \dfrac{3}{2}$.