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