The zeros for the given equation are A.
![x = {2 \pm √(13)}.](https://img.qammunity.org/2021/formulas/mathematics/middle-school/yd5xtywqjn4cvnksuvzqlhk0fw4xl7lw32.png)
Explanation:
Step 1:
To solve for x in a polynomial equation, we use the formula
![x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a}.](https://img.qammunity.org/2021/formulas/mathematics/middle-school/jojzh87cttz1ykofsul3wf115am6rnjp1f.png)
Here a is the coefficient of
, b is the coefficient of
and c is the coefficient of the constant term.
In the given equation,
.
Step 2:
Substituting the values of a, b, and c in the equation, we get
![x=\frac{-b \pm \sqrt{b^(2)-4 a c}}{2 a} =\frac{-(-4) \pm \sqrt{(-4)^(2)-4 (1)(-9)}}{2 (1)} = (4 \pm √((16+36))/(2 ).](https://img.qammunity.org/2021/formulas/mathematics/middle-school/y79j32daxd0v7h5x7rww2yhh3f3x9z92c8.png)
![(4 \pm √((16+36))/(2 ) = (4 \pm √((52))/(2 ) = (4 \pm2 √((13))/(2 )= {2 \pm √(13)}.](https://img.qammunity.org/2021/formulas/mathematics/middle-school/dth8nhpdr47qyb2e8i1tqssqoi8f4rqply.png)
So the answer is option A.
![x = {2 \pm √(13)}.](https://img.qammunity.org/2021/formulas/mathematics/middle-school/yd5xtywqjn4cvnksuvzqlhk0fw4xl7lw32.png)