How do you find the zeros, real and imaginary, of #y=3x^2+17x-43# using the quadratic formula?

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Answer 1 · 2016-07-18T06:42:48

We have real zeros #(-17-sqrt805)/6# or #(-17+sqrt805)/6#

Zeros of #y=3x^2+17x-43# are those values of #x# for which #y=0#. Hence, zeros are given by the solution of quadratic equation #3x^2+17x-43=0#.

Quadratic formula gives solution of equation #ax^2+bx+c=0# as #x=(-b+-sqrt(b^2-4ac))/(2a)#. Hence for #3x^2+17x-43=0#, we have

#x=(-17+-sqrt(17^2-4*3*(-43)))/(2*3)#

= #(-17+-sqrt(289+516))/6#

= #(-17+-sqrt805)/6#

= #(-17-sqrt805)/6# or #(-17+sqrt805)/6#,

Which are real zeros of #3x^2+17x-43=0#.