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]}

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