Answer:
Given the equation: \(3x^2+14x+16=0\)
The general equation of quadratic formula for \(ax^2+bx+c=0\) where a, b and c are constant;
then, the solution for this equation is given by;
\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) ......[1]
From the equation, \(3x^2+14x+16=0\) we have
a =2, b=14 and c=16
Substitute these values in [1] to solve for x;
\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-14\pm\sqrt{(14)^2-4(3)(16)}}{2(3)}\)
or
\(x=\frac{-14\pm\sqrt{196-192}}{6} =\frac{-14\pm\sqrt{4}}{6}\)
Simplify:
\(x=\frac{-14\pm\sqrt{4}}{6} =\frac{-14\pm 2}{6}\)
then,
\(x=\frac{-14+2}{6} =\frac{-12}{6} =-2\) and
\(x=\frac{-14-2}{6} =\frac{-16}{6} =\frac{-8}{3}\)
Therefore, the solutions for the given equation is; x =-2 and \(x =-\frac{8}{3}\)
