What are the solutions to the equation roots (s+1)^2+4 ?

The solutions to the equation roots (s+1)^2+4 are s=2i-1,s=-2i-1

Find the zeros of roots (s+1)^2+4

The zeros of roots (s+1)^2+4 are s=2i-1,s=-2i-1

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roots

(s + 1)^{2} + 4

s = 2 i - 1, s = - 2 i - 1

Solution steps

(s + 1)^{2} + 4

The roots are the intercepts with the x-axis

(y = 0)

(s + 1)^{2} + 4 = 0

Move

4

to the right side

(s + 1)^{2} = - 4

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

s + 1 = - 4 : s = 2 i - 1

Solve

s + 1 = - - 4 : s = - 2 i - 1

The solutions to the quadratic equation are:

s = 2 i - 1, s = - 2 i - 1

9 (x + 9)^{2} - 36 = 0

(2 x - 1) (3 x + 1) = 0

12 x^{2} - 24 x - 4 = 0

2 x^{2} - 8 x - 15 = 0

4 x + 6 x (x - 5) + 12 = x^{2} + 7

What are the solutions to the equation roots (s+1)^2+4 ?
The solutions to the equation roots (s+1)^2+4 are s=2i-1,s=-2i-1
Find the zeros of roots (s+1)^2+4
The zeros of roots (s+1)^2+4 are s=2i-1,s=-2i-1