Answer :
[tex]\begin{gathered} \text{When we have a polynomial } \\ ax^2+bx+c=0 \\ \text{The roots of the polynomial are given by} \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ \text{ Thus the zeros of } \\ f(x)=2x^2+x-28\text{ } \\ \text{are} \\ \\ \\ x=\frac{-1\pm\sqrt[]{1^2-4(2)(-28)}}{2(2)} \\ x=\frac{-1\pm\sqrt[]{1+224}}{4} \\ \\ x=\frac{-1\pm\sqrt[]{225}}{4} \\ x=\frac{-1\pm15}{4} \\ \\ we\text{ have two roots} \\ x_1=\frac{-1-15}{4}=-\frac{16}{4}=-4 \\ \\ x_2=\frac{-1+15}{4}=\frac{14}{4}=\frac{7}{2}=3.5 \end{gathered}[/tex]