You have the following function:
![f(x)=x^2+5x+6](https://img.qammunity.org/2023/formulas/mathematics/college/95tjsezcbon9xy8vypbkmdte7clzwyozxo.png)
in order to find the zeros of the previous function, use the quadratic formula:
![x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/rxvf73usjbbwyik14knxdemoz21vfz2ufc.png)
where a, b and c are the coefficients of the polynomial. In this case:
a = 1
b = 5
c = 6
replace the previous values of the parameters into the formula for x:
![\begin{gathered} x=\frac{-5\pm\sqrt[]{5^2-4(1)(6)}}{2(1)} \\ x=\frac{-5\pm\sqrt[]{25-24}}{2}=(-5\pm1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u68mrfiklbm8g2xh5717b3e2asgaip3kki.png)
hence the solution for x are:
x = (-5-1)/2 = -6/2 = -3
x = (-5+1)/2 = -4/2 = -2
A) x = -2 , -3