What are the solutions to the equation 4(x+1)^2+12=0 ?

The solutions to the equation 4(x+1)^2+12=0 are x=sqrt(3)i-1,x=-sqrt(3)i-1

Find the zeros of 4(x+1)^2+12=0

The zeros of 4(x+1)^2+12=0 are x=sqrt(3)i-1,x=-sqrt(3)i-1

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4 (x + 1)^{2} + 12 = 0

x = 3 i - 1, x = - 3 i - 1

Solution steps

4 (x + 1)^{2} + 12 = 0

Move

12

to the right side

4 (x + 1)^{2} = - 12

Divide both sides by

4

(x + 1)^{2} = - 3

For

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

the solutions are

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

Solve

x + 1 = - 3 : x = 3 i - 1

Solve

x + 1 = - - 3 : x = - 3 i - 1

The solutions to the quadratic equation are:

x = 3 i - 1, x = - 3 i - 1

7 v^{2} - 14 v + 4 = 3

9 x^{2} + 30 x + 28 = 0

3 x^{2} + 15 x = 72

x^{2} - 18 x + 79 = 0

(x - 2) (x - 9) = 0

What are the solutions to the equation 4(x+1)^2+12=0 ?
The solutions to the equation 4(x+1)^2+12=0 are x=sqrt(3)i-1,x=-sqrt(3)i-1
Find the zeros of 4(x+1)^2+12=0
The zeros of 4(x+1)^2+12=0 are x=sqrt(3)i-1,x=-sqrt(3)i-1