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

Question

1873 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-17T18:14:19

$\frac{- 35 \pm \sqrt{1345}}{6}$ $(-35+-sqrt(1345))/6$

$- \frac{1}{5} x 2 - \frac{7}{3} x + \frac{2}{3} = 0$ $−1/5x2−7/3x+2/3=0$

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

$- \frac{3}{15} x^{2} - \frac{35}{15} x + \frac{10}{15} = 0$ $−3/15x^2−35/15x+10/15=0$

Now multiply both side by 15:

$- 3 x^{2} - 35 x + 10 = 0$ $−3x^2−35x+10=0$

now apply the second-degree formula:

$\frac{35 \pm \sqrt{35^{2} - 4 (- 3) (10)}}{2 \cdot (- 3)}$ $(35+-sqrt(35^2-4(-3)(10)))/(2*(-3))$

$\frac{35 \pm \sqrt{1225 + 120}}{- 6}$ $(35+-sqrt(1225+120))/(-6)$

$\frac{- 35 \pm \sqrt{1345}}{6}$ $(-35+-sqrt(1345))/6$

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=−15x2−73x+23 y = -1/5x^2 - 7/3x + 2/3 using the quadratic formula?