What are the solutions to the equation 4(x+2)^2+13=29 ?

The solutions to the equation 4(x+2)^2+13=29 are x=0,x=-4

Find the zeros of 4(x+2)^2+13=29

The zeros of 4(x+2)^2+13=29 are x=0,x=-4

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4 (x + 2)^{2} + 13 = 29

x = 0, x = - 4

Solution steps

4 (x + 2)^{2} + 13 = 29

Move

13

to the right side

4 (x + 2)^{2} = 16

Divide both sides by

4

(x + 2)^{2} = 4

For

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

the solutions are

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

Solve

x + 2 = 4 : x = 0

Solve

x + 2 = - 4 : x = - 4

The solutions to the quadratic equation are:

x = 0, x = - 4

y, x^{2} + y^{2} = x y

y, 3 x^{2} + 5 x^{2} y^{2} = 2 y

x^{2} + 2 x - 4 = - 4

4 x^{2} + 5 x = 4 x

3 y^{2} - 24 y - 47 = 0

What are the solutions to the equation 4(x+2)^2+13=29 ?
The solutions to the equation 4(x+2)^2+13=29 are x=0,x=-4
Find the zeros of 4(x+2)^2+13=29
The zeros of 4(x+2)^2+13=29 are x=0,x=-4