How do you find the zeros, real and imaginary, of #y= x^2-3x-16# using the quadratic formula?

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Answer 1 · 2017-12-19T23:15:53

#x_1=(3+sqrt73)/2# and #x_2=(3-sqrt73)/2#

#Delta=(-3)^2-4*1*(-16)=73#

Hence solution of y, #x_1=(3+sqrt73)/2# and #x_2=(3-sqrt73)/2#

Answer 2 · 2017-12-19T23:30:18

#x = (3 +- sqrt(73))/2#

The quadratic formula is #x = (-b+- sqrt(b^2 - 4ac))/(2a)#

Your equation #y = x^2 - 3x - 16# is in standard quadratic form, or in the form of #ax^2 + bx + c#

So...
#a = 1#
#b = -3#
#c = -16#

Now let's plug them in the quadratic formula:
#x = (-(-3) +- sqrt((-3)^2 - 4(1)(-16)))/(2(1))#
#x = (3 +- sqrt(9 + 64))/2#
#x = (3 +- sqrt(73))/2#