GIven,
$$
\begin{aligned}
&f(x)=6 x^{2}-3-7 x \\
&\text { We put } f(x)=0 \\
&=6 x^{2}-3-7 x=0 \\
&=6 x^{2}-9 x+2 x-3=0 \\
&=3 x(2 x-3)+1(2 x-3)=0 \\
&=(2 x-3)(3 x+1)=0
\end{aligned}
$$
This gives us 2 zeros, for
$$
x=\frac{3}{2} \text { and } x=\frac{-1}{3}
$$
Hence, the zeros of the quadratic equation are \(\frac{3}{2}\) and \(\frac{-1}{3}\).
Now, for verification Sum of zeros \(=\frac{\operatorname{coc} f f i d i e n t ~ o f ~}{\cos f f \text { icicnt of } x^{2}}\)
\(\frac{3}{2}+\left(\frac{-1}{3}\right)=-\frac{(-7)}{6}\) \(\frac{7}{6}=\frac{7}{6}\) Product of roots \(=\frac{\text { comstant }}{\text { ceffident of } x^{2}}\) \(\frac{3}{2} \times\left(\frac{-1}{3}\right)=\frac{(-3)}{6}\) \(\frac{-1}{2}=\frac{-1}{2}\)
Therefore, the relationship between zeros and their coefficlents Is verifled.
