How do you find the roots, real and imaginary, of $y= 12x^2+3x-6$ using the quadratic formula?

Question

How do you find the roots, real and imaginary, of $y= 12x^2+3x-6$ using the quadratic formula?

1 Answer

Impact of this question

1354 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-15T23:54:56

$x = -1/8+- sqrt33/8$

Explanation:

$y = 12x^2 + 3x - 6 = 4x^2 + x - 2 = 0$
$D = d^2 = b^2 - 4ac = 1 + 32 = 33$ --> $d = +-sqrt33$
There are 2 real roots:
$x = -b/(2a) +- d/(2a) = -1/8 +- sqrt33/8$

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of $y= 12x^2+3x-6$ using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the roots, real and imaginary, of y= 12x^2+3x-6 y= 12x^2+3x-6 using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the roots, real and imaginary, of $y= 12x^2+3x-6$ using the quadratic formula?