Answer:
The correct option is A) x = -8/3, –2.
Step-by-step explanation:
Consider the provided equation \(3x^2 + 14x + 16 = 0\)
The above equation is in the form of \(ax^2 +bx + c = 0\)
The quadratic formula to find the root of the equation is:
\({\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\ \ }\)
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:
\({\displaystyle x={\frac {-14\pm {\sqrt {14^{2}-4(3)(16)}}}{2(3)}}\ \ }\)
\({\displaystyle x={\frac {-14\pm {\sqrt {196-192}}}{6}}\ \ }\)
\({\displaystyle x={\frac {-14\pm {\sqrt {4}}}{6}}\ \ }\)
\({\displaystyle x={\frac {-14\pm2}{6}}\ \ }\)
\({\displaystyle x={\frac {-14+2}{6}}\ \ } or\ {\displaystyle x={\frac {-14-2}{6}}\ \ }\)
\({\displaystyle x={\frac {-12}{6}}\ \ } or\ {\displaystyle x={\frac {-16}{6}}\ \ }\)
\({\displaystyle x=-2}\ \ } or\ {\displaystyle x={\frac {-8}{3}}\ \ }\)
Hence the solutions of the provided equation is \({\displaystyle x=-2}\ \ } or\ {\displaystyle x={\frac {-8}{3}}\ \ }\).
Thus, the correct option is A) x = -8/3, –2.
