How do you find the zeros, real and imaginary, of #y= x^2-5x+6# using the quadratic formula?

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Answer 1 · 2018-06-18T00:33:54

#x=3,# #2#

#y=x^2-5x+6# is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=1#, #b=-5#, #c=6#

The zeroes are the values for #x# when #y=0#.

Substitute #0# for #y#.

#0=x^2-5x+6#

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

Plug in the known values and solve.

#x=(-(-5)+-sqrt((-5)^2-4*1*6))/(2*1)#

#x=(5+-sqrt1)/2#

Simplify.

#x=(5+-1)/2#

#x=(5+1)/2#, #(5-1)/2#

#x=6/2#, #4/2#

Simplify.

#x=3,# #2#

graph{y=x^2-5x+6 [-10, 10, -5, 5]}