Question

What are the solutions of 3x2 + 14x + 16 = 0?

Answer 1

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 $x =-\frac{8}{3}$

Answer 2

The correct answer is
a) x=-8/3,-2