How do you find the zeros, real and imaginary, of $y=-x^2+4x-42$ using the quadratic formula?

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Answer 1 · 2016-06-07T16:25:17

The roots are $x_1=2-sqrt(38)i$ and $x_2=2+sqrt(38)i$ .

Function $y=ax^2+bx+c$ has two roots (zeros) $x_{1,2}=(-b+-sqrt(b^2-4ac))/(2a)$ or one double root if $b^2=4ac$ .

In this case $a=-1,b=4,c=-42$

So you can blindly plug in those values:

$x_{1,2}=(-(4)+-sqrt((4)^2-4(-1)(-42)))/(2(-1))=(-4+-sqrt(16-168))/-2 =(4+-sqrt(-152))/2=(4+-2sqrt(38)i)/2=2+-sqrt(38)i$

The roots are $x_1=2-sqrt(38)i$ and $x_2=2+sqrt(38)i$ .

$sqrt(38)=6.164414...$

How do you find the zeros, real and imaginary, of y=-x^2+4x-42y=-x^2+4x-42 using the quadratic formula?