What are the solutions to the equation 0=(x+3)^2-4 ?

The solutions to the equation 0=(x+3)^2-4 are x=-1,x=-5

Find the zeros of 0=(x+3)^2-4

The zeros of 0=(x+3)^2-4 are x=-1,x=-5

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0 = (x + 3)^{2} - 4

x = - 1, x = - 5

Solution steps

0 = (x + 3)^{2} - 4

Switch sides

(x + 3)^{2} - 4 = 0

Move

4

to the right side

(x + 3)^{2} = 4

For

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

the solutions are

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

Solve

x + 3 = 4 : x = - 1

Solve

x + 3 = - 4 : x = - 5

The solutions to the quadratic equation are:

x = - 1, x = - 5

2 x^{2} + 7 x - 4 \leq 0

d^{2} - 12 d + 36

x^{2} + 16 x - 2 = 0

\frac{2 x ^{2} + 2 x}{3 x ^{2} + 3 x}

2 x^{2} + 6 x - 36

What are the solutions to the equation 0=(x+3)^2-4 ?
The solutions to the equation 0=(x+3)^2-4 are x=-1,x=-5
Find the zeros of 0=(x+3)^2-4
The zeros of 0=(x+3)^2-4 are x=-1,x=-5