Explanation
The zero of a function is any replacement for the variable that will produce an answer of zero,so to find the zeros of this function we need set the equation = 0, and then solve for x
[tex]2x^2+7x+3=0[/tex]The rigth side of the equation is zero, so we can solve
[tex]\begin{gathered} 2x^2+7x+3=0\Rightarrow ax^2+bx+c \\ \end{gathered}[/tex]we can use the quadratic formula
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{replace} \\ x=\frac{-7\pm\sqrt[]{7^2-4\cdot2\cdot3}}{2\cdot2} \\ x=\frac{-7\pm\sqrt[]{49-24}}{4} \\ x=\frac{-7\pm\sqrt[]{25}}{4} \\ x=\frac{-7\pm5}{4} \end{gathered}[/tex]therefore.
[tex]\begin{gathered} x_1=\frac{-7+5}{4}=-\frac{2}{4}=-\frac{1}{2} \\ x_2=\frac{-7-5}{4}=-\frac{12}{4}=-3 \end{gathered}[/tex]so, the answer is
[tex]b)-3,\text{ -}\frac{1}{2}[/tex]I hope this helps you
