To find the zeros of a function f(x), we set f(x)=0 and then solve for x
Here. we have
[tex]\begin{gathered} f(x)=x+\frac{3}{2}x^2-6 \\ \text{ setting }f(x)=0 \\ \Rightarrow\frac{3}{2}x^2+x-6=0 \\ \text{ multiplying each term by 2} \\ \Rightarrow2(\frac{3}{2}x^2)+2(x)-2(6)=0 \\ \Rightarrow3x^2+2x-12=0 \\ \Rightarrow3x^2+2x=12 \\ \text{ Dividing each term by 3} \\ \Rightarrow x^2+\frac{2}{3}x=4 \\ \text{ Adding (}\frac{1}{2}\times\frac{2}{3})^2\text{ =}\frac{1}{9}\text{to both sides;} \\ \Rightarrow x^2+\frac{2}{3}x+\frac{1}{9}=4+\frac{1}{9} \\ \Rightarrow(x+\frac{1}{3})^2=\frac{37}{9} \\ \Rightarrow(x+\frac{1}{3})=\frac{\pm\sqrt[]{37}}{3} \\ \Rightarrow x=-\frac{1}{3}\pm\frac{\sqrt[]{37}}{3} \\ \text{ The zeros of f(x) is given as;} \\ \Rightarrow x=-\frac{1}{3}+\frac{\sqrt[]{37}}{3}\text{ or }x=-\frac{1}{3}-\frac{\sqrt[]{37}}{3} \end{gathered}[/tex]