Answer:
Option b)
.
Explanation:
We have to solve the equation x²+6x+4 =0 for the value of x.
As we know from any quadratic equation ax²+bx+c=0
the value of
![x=\frac{-b\pm \sqrt{b^(2)-4ac}}{2a}](https://img.qammunity.org/2020/formulas/mathematics/high-school/abg2ze6szl4l3ajsl23tkzpysujdbdj2dc.png)
Now with the help of this formula we will solve the equation
x²+6x+4 = 0
By putting the values of a= 1, b=6, c=4 in the formula
![x=\frac{-6\pm \sqrt{6^(2)-4* 1* 4}}{2* 1}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9cdz0fohxw4ey5zgh37m6wpdiuiev5d3he.png)
![x=(-6\pm √(36-16))/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/en6wi2834wlgqrsba5b49e0itwh97vkd9q.png)
![x=(-6\pm √(20))/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hy469epupldxsynany3jbct8xlabygzmho.png)
![x=(-6\pm √(4* 5))/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/33si6aop82cdi5q6u4kvlrrqnaipgiktpn.png)
![x=(-6\pm 2√(5))/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hdwjp8farwui91ryu5kb6ecsh7cg9vobui.png)
![x=(2(-3\pm √(5)))/(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/rbgmdxpre5znlmwjdywgdugw1a6y740url.png)
![x=-3\pm √(5)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bsac09rh1ti3b1v4g5tlhunp4zzk3q5odo.png)
so the final answer is
![x=(-3\pm √(5))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/f2vrja5nmk4s75dee8cd81mjtd09i4vq0j.png)