Question

find the zeros of 6x square -7x -3 and vertify the difference between the zeros and the coefficient​

Answer 1

Step-by-step explanation:

$6{x}^{2} - 7x - 3 = 0 \\ \alpha + \beta = - \frac{ - 7}{6} \\ \alpha \beta = \frac{ - 3}{6}$