What are the solutions to the equation -2(x+1)^2-5=-55 ?

The solutions to the equation -2(x+1)^2-5=-55 are x=4,x=-6

Find the zeros of -2(x+1)^2-5=-55

The zeros of -2(x+1)^2-5=-55 are x=4,x=-6

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- 2 (x + 1)^{2} - 5 = - 55

x = 4, x = - 6

Solution steps

- 2 (x + 1)^{2} - 5 = - 55

Move

5

to the right side

- 2 (x + 1)^{2} = - 50

Divide both sides by

- 2

(x + 1)^{2} = 25

For

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

the solutions are

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

Solve

x + 1 = 25 : x = 4

Solve

x + 1 = - 25 : x = - 6

The solutions to the quadratic equation are:

x = 4, x = - 6

x, x^{2} + 3 x + h = 9 + h

(x + 9) (2 x + 1) = 100

6 x^{2} - 24 x = 30

0 = - 16 t^{2} + 12 t + 120

(x + 1) (2 x + 3) (3 x + 4) = (6 x^{2} + 5) (x - 3)

What are the solutions to the equation -2(x+1)^2-5=-55 ?
The solutions to the equation -2(x+1)^2-5=-55 are x=4,x=-6
Find the zeros of -2(x+1)^2-5=-55
The zeros of -2(x+1)^2-5=-55 are x=4,x=-6