How do you find the roots, real and imaginary, of $y= x^2 - 8x -9- (2x-1)^2$ using the quadratic formula?

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Answer 1 · 2017-12-22T09:38:26

see below

$x^2−8x−9−(2x−1)^2$

$=>x^2−8x−9−(4x^2-2x+1)$

$=>x^2−8x−9−4x^2+2x-1)$

$=>-3x^2−6x−10$

$=>3x^2+6x+10$

$D=b^2-4acquad=>quad36-4*3*10$
$D<0=36-120=-84$

$x_(1,2)=(-b+-sqrtD)/(2a)=(-6+-sqrt(7*3*2*2*i^2))/(6)$

$x_(1,2)=(-6+-2isqrt(7*3))/(6)=(cancel2(-3+-isqrt(21)))/(cancel2*3)$

$x_(1)=(-3-isqrt(21))/(3)=-1-(isqrt(21))/3$

$x_(2)=(-3+isqrt(21))/(3)=-1+(isqrt(21))/3$

How do you find the roots, real and imaginary, of y= x^2 - 8x -9- (2x-1)^2 y= x^2 - 8x -9- (2x-1)^2 using the quadratic formula?