Answer
$\left\{\dfrac{-1-\sqrt{13}}{2}, \dfrac{-1+\sqrt{13}}{2}\right\}$
Work Step by Step
To find the zeros of the given function, set $f(x)=0$ then solve for $x$ using the Quadratic Formula:
$$0=x^2+x-3$$
In this equation, we have: $a=1, b=1, \ c=-3$
Find the zeros using the Quadratic Formula to obtain:
$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$.
$x=\dfrac{-1\pm\sqrt{1^2-4(1)(-3)}}{2(1)}\\
=\dfrac{-1\pm\sqrt{1+12}}{2}\\=\dfrac{-1\pm\sqrt{13}}{2}\\$
Thus, the zeros are:
$\left\{\dfrac{-1-\sqrt{13}}{2}, \dfrac{-1+\sqrt{13}}{2}\right\}$
