If the equation x^2+ (y+1)x -1 =0 has equal roots what are the values of y

We'll try to calculate the value of y using
Viete's relations:

x1 + x2 =
-b/a

x1*x2 = c/a

We'll
identify the coefficients of the equation:

a =
1

b = y+1

c =
-1

We also know, from enunciation, taht the roots are
equal:x1=x2

x1 + x2 = 2x = - y - 1
(1)

x^2 = -1 (2)

We'll raise
to square (1):

4x^2 =
(y+1)^2

But x^2 = -1. We'll substitute x^2 by the value -1
and we'll expand the square from the right side:

-4 = y^2 +
2y + 1

We'll add 4 both sides and we'll use symmetric
property:

y^2 + 2y + 1 + 4 =
0

y^2 + 2y + 5 = 0

We'll apply
the quadratic formula:

y1 =
[-2+sqrt(4-20)]/2

y1 = [-2 +
sqrt(-16)]/2

y1 =
(-2+4i)/2

y1 = -1 +
2i

y2 = -1 -
2i