How do you find the zeros of $y = -1/5x^2 - 7/3x + 2/3$ using the quadratic formula?

Question

1890 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-17T18:14:19

$(-35+-sqrt(1345))/6$

$−1/5x2−7/3x+2/3=0$

The first thing to do is to simplify the equation replacing the reaction by the equivalent fraction with the same denominator:

$−3/15x^2−35/15x+10/15=0$

Now multiply both side by 15:

$−3x^2−35x+10=0$

now apply the second-degree formula:

$(35+-sqrt(35^2-4(-3)(10)))/(2*(-3))$

$(35+-sqrt(1225+120))/(-6)$

$(-35+-sqrt(1345))/6$

Theres is now way to simplifiy sqrt(1345) so the solution may stay lique this.

In the case of the need of a number the solution are:

0.27904 and -11.9457

How do you find the zeros of y = -1/5x^2 - 7/3x + 2/3 y = -1/5x^2 - 7/3x + 2/3 using the quadratic formula?