What are the solutions to the equation 2(x+1)^2+7=1 ?

The solutions to the equation 2(x+1)^2+7=1 are x=sqrt(3)i-1,x=-sqrt(3)i-1

Find the zeros of 2(x+1)^2+7=1

The zeros of 2(x+1)^2+7=1 are x=sqrt(3)i-1,x=-sqrt(3)i-1

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2 (x + 1)^{2} + 7 = 1

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

Solution steps

2 (x + 1)^{2} + 7 = 1

Move

7

to the right side

2 (x + 1)^{2} = - 6

Divide both sides by

2

(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

m^{2} + 4 m + 8 = 0

x, (2 x + j^{3} y)^{2} - 6 - j^{12} = 0

0 = x^{2} - 2 x + 6

x^{2} - 20 x + 72 = - 3 x - 3 x

x, 2 x^{2} + 5 x + 3 = 0

What are the solutions to the equation 2(x+1)^2+7=1 ?
The solutions to the equation 2(x+1)^2+7=1 are x=sqrt(3)i-1,x=-sqrt(3)i-1
Find the zeros of 2(x+1)^2+7=1
The zeros of 2(x+1)^2+7=1 are x=sqrt(3)i-1,x=-sqrt(3)i-1