How do you find the zeros, real and imaginary, of $y= 3x^2+4x+2$ using the quadratic formula?

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Answer 1 · 2015-12-21T14:03:45

Substitute the coefficients into the equation and evaluate.

The quadratic equation is $x=\frac{-b\pm\sqrt{b^2-4ac\ }}{2a}$
In the example a = 3, b = 4 and c = 2

$x = (-4 +- sqrt(4^2 - 4*3*2))/(2*3)$
$x= (-4 +- sqrt(16 - 24))/6$
$x = -4/6 +- sqrt(-8)/6$
$x = -2/3(1 +sqrt(-2))$ or $-2/3(1 - sqrt(-2))$

The equation only has imaginary roots. The graph never touches or crosses the x axis.

How do you find the zeros, real and imaginary, of y= 3x^2+4x+2 y= 3x^2+4x+2 using the quadratic formula?