How do you find the zeros, real and imaginary, of $y=- x^2-22x+6$ using the quadratic formula?

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Answer 1 · 2016-04-10T10:59:06

Zeros are $-11+sqrt127$ and $-11-sqrt127$

The roots of general form of equation $ax^2+bx+c=0$ are zeros of the general form of equation $y=ax^2+bx+c$ .

In the given equation $y=-x^2-22x+6$ , we see that $a=-1$ , $b=-22$ and $c=6$ .

As discriminant $b^2-4ac=(-22)^2-4(-1)(6)=484+24=508$

As discriminant is positive but not a complete square, we have real but irrational roots.

Hence, using quadratic formula $(-b+-sqrt(b^2-4ac))/(2a)$ ,

the zeros are the given equation are $x=(-(-22)+-sqrt508)/(2*(-1))$

or $x=-(22+-sqrt508)/2=-(22+-2sqrt127)/2$ i.e.

zeros are $-11+sqrt127$ and $-11-sqrt127$

How do you find the zeros, real and imaginary, of y=- x^2-22x+6 y=- x^2-22x+6 using the quadratic formula?