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

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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