Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients: $6x^{2}-3-7x$.

(N/A) First,rewrite the polynomial in standard form: $6x^{2}-7x-3$.
To find the zeroes,factorize the quadratic polynomial:
$6x^{2}-7x-3 = 6x^{2}-9x+2x-3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)$.
The value of $6x^{2}-7x-3$ is zero when $3x+1=0$ or $2x-3=0$,which gives $x = -1/3$ or $x = 3/2$.
Therefore,the zeroes are $\alpha = -1/3$ and $\beta = 3/2$.
Verification of the relationship between zeroes and coefficients:
Sum of zeroes: $\alpha + \beta = -1/3 + 3/2 = (-2+9)/6 = 7/6 = -(-7)/6 = -(\text{Coefficient of } x) / (\text{Coefficient of } x^{2})$.
Product of zeroes: $\alpha \times \beta = (-1/3) \times (3/2) = -3/6 = -1/2 = -3/6 = (\text{Constant term}) / (\text{Coefficient of } x^{2})$.