How do you find the zeros, real and imaginary, of $y=x^2+32x+44$ using the quadratic formula?

Question

How do you find the zeros, real and imaginary, of $y=x^2+32x+44$ using the quadratic formula?

1 Answer

Impact of this question

1522 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-24T08:48:08

Substitute the coefficients into the quadratic formula to find:

$x=-16+-2sqrt(53)$

Explanation:

$x^2+32x+44$ is of the form $ax^2+bx+c$
with $a=1$ , $b=32$ and $c=44$ .

This has zeros given by the quadratic formula:

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

$=(-32+-sqrt(32^2-(4xx1xx44)))/(2*1)$

$=(-32+-sqrt(1024-176))/2$

$=(-32+-sqrt(848))/2$

$=(-32+-sqrt(4^2*53))/2$

$=(-32+-4sqrt(53))/2$

$=-16+-2sqrt(53)$

Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of $y=x^2+32x+44$ using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the zeros, real and imaginary, of y=x^2+32x+44y=x^2+32x+44 using the quadratic formula?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the zeros, real and imaginary, of $y=x^2+32x+44$ using the quadratic formula?