How do you find the zeros, real and imaginary, of #y=5x^2-x-2# using the quadratic formula?

1 Answer
Prof_S
Jan 11, 2016

The zeroes are given by y=#(1 (+/-) sqrt(41))/10#

Explanation:

The quadratic equation, which gives the roots of a second degree polynomial of the form #ax^2+bx+c#, is:

#(-b (+/-) sqrt(b^2 - 4ac))/(2a)#.

Substituting for a the value 5, for b the value -1, and for c the value -2, gives #(-(-1) (+/-) sqrt((-1)^2 - 4(5)(-2)))/(2(5))#. This gives the stated values.

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