What are the solutions to the equation 21=(x+3)^2-4 ?

The solutions to the equation 21=(x+3)^2-4 are x=2,x=-8

Find the zeros of 21=(x+3)^2-4

The zeros of 21=(x+3)^2-4 are x=2,x=-8

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

x = 2, x = - 8

Solution steps

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

Switch sides

(x + 3)^{2} - 4 = 21

Move

4

to the right side

(x + 3)^{2} = 25

For

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

the solutions are

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

Solve

x + 3 = 25 : x = 2

Solve

x + 3 = - 25 : x = - 8

The solutions to the quadratic equation are:

x = 2, x = - 8

x^{2} + 16 x = 22

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

(1 - x) (2 - x) = 0

4 y^{2} - 1.6 y + 0.1225 = 0

f, 2 f + x (f)^{2} = 10

What are the solutions to the equation 21=(x+3)^2-4 ?
The solutions to the equation 21=(x+3)^2-4 are x=2,x=-8
Find the zeros of 21=(x+3)^2-4
The zeros of 21=(x+3)^2-4 are x=2,x=-8