How do you find the roots, real and imaginary, of $y=3x^2+5x+2$ using the quadratic formula?

Question

2266 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-12T00:09:50

Substitute the coefficients into the quadratic formula to find zeros:

$x=-1$ and $x=-2/3$

$y=3x^2+5x+2$ is of the form $ax^2+bx+c$ with $a=3$ , $b=5$ and $c=2$

It has zeros given by the quadratic formula:

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$=(-5+-sqrt(5^2-(4*3*2)))/(2*3)$

$=(-5+-sqrt(25-24))/6$

$=(-5+-1)/6$

That is $x=-1$ and $x=-2/3$

How do you find the roots, real and imaginary, of y=3x^2+5x+2y=3x^2+5x+2 using the quadratic formula?