How do you find the roots, real and imaginary, of $y=-(x -2)^2-4x+5$ using the quadratic formula?

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Answer 1 · 2017-07-13T06:05:18

${-1,1}$

We have roots of the equation and zeros of the function.

Here $y=-(x-2)^2-4x+5$

= $-x^2+4x-4-4x+5$

= $-x^2+1$

According ro quadratic formula, zeros of function $f(x)=ax^2+bx+c$ are $(-b+-sqrt(b^2-4ac))/(2a)$

Here $a=-1$ , $b=0$ and $c=1$

hence zeros are $(+-sqrt(-4xx(-1)xx1))/(-2)=+-2/(-2)$ or $1,-1$

How do you find the roots, real and imaginary, of y=-(x -2)^2-4x+5y=-(x -2)^2-4x+5 using the quadratic formula?