What are the solutions to the equation 3(x+1)^2=-240 ?

The solutions to the equation 3(x+1)^2=-240 are x=4sqrt(5)i-1,x=-4sqrt(5)i-1

Find the zeros of 3(x+1)^2=-240

The zeros of 3(x+1)^2=-240 are x=4sqrt(5)i-1,x=-4sqrt(5)i-1

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3 (x + 1)^{2} = - 240

x = 45 i - 1, x = - 45 i - 1

Solution steps

3 (x + 1)^{2} = - 240

Divide both sides by

3

(x + 1)^{2} = - 80

For

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

the solutions are

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

Solve

x + 1 = - 80 : x = 45 i - 1

Solve

x + 1 = - - 80 : x = - 45 i - 1

The solutions to the quadratic equation are:

x = 45 i - 1, x = - 45 i - 1

15 b^{2} - 8 b - 3 = - 4

9 x^{2} + 6 x - 63 = 0

x^{2} - 2 x 2 - 5 = 0

2 W^{2} + 3 W - 54 = 0

k^{2} + 10 k + 15 = 0

What are the solutions to the equation 3(x+1)^2=-240 ?
The solutions to the equation 3(x+1)^2=-240 are x=4sqrt(5)i-1,x=-4sqrt(5)i-1
Find the zeros of 3(x+1)^2=-240
The zeros of 3(x+1)^2=-240 are x=4sqrt(5)i-1,x=-4sqrt(5)i-1