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

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

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

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

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

x = 1, x = - 3

Solution steps

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

Switch sides

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

Move

4

to the right side

(x + 1)^{2} = 4

For

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

the solutions are

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

Solve

x + 1 = 4 : x = 1

Solve

x + 1 = - 4 : x = - 3

The solutions to the quadratic equation are:

x = 1, x = - 3

\frac{3}{4} x^{2} + 1460 x + 300 = - \frac{1}{4} x^{2} + 1500 x

x, x^{2} + 8 x + 16 = 0

x^{2} + 7 = 10

9 e^{2} + 8 e = - 3

x^{2} + 5 x = 5

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