The correct option is A) x = -8/3, β2.
Step-by-step explanation:
Consider the provided equation ![3x^2 + 14x + 16 = 0](/tpl/images/0482/0507/afccb.png)
The above equation is in the form of ![ax^2 +bx + c = 0](/tpl/images/0482/0507/559b3.png)
The quadratic formula to find the root of the equation is:
![{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ \ }](/tpl/images/0482/0507/5d660.png)
By comparing the above equation with the general equation we can conclude that:
a = 3, b = 14, and c = 16
Substitute the respective values in the above formula:
![{\displaystyle x={\frac {-14\pm {\sqrt {14^{2}-4(3)(16)}}}{2(3)}}\ \ }](/tpl/images/0482/0507/e4eb8.png)
![{\displaystyle x={\frac {-14\pm {\sqrt {196-192}}}{6}}\ \ }](/tpl/images/0482/0507/82845.png)
![{\displaystyle x={\frac {-14\pm {\sqrt {4}}}{6}}\ \ }](/tpl/images/0482/0507/8f5cb.png)
![{\displaystyle x={\frac {-14\pm2}{6}}\ \ }](/tpl/images/0482/0507/4ed05.png)
![{\displaystyle x={\frac {-14+2}{6}}\ \ } or\ {\displaystyle x={\frac {-14-2}{6}}\ \ }](/tpl/images/0482/0507/cc50d.png)
![{\displaystyle x={\frac {-12}{6}}\ \ } or\ {\displaystyle x={\frac {-16}{6}}\ \ }](/tpl/images/0482/0507/a3e87.png)
![{\displaystyle x=-2}\ \ } or\ {\displaystyle x={\frac {-8}{3}}\ \ }](/tpl/images/0482/0507/6a603.png)
Hence the solutions of the provided equation is
.
Thus, the correct option is A) x = -8/3, β2.