What are the solutions to the equation 4(x+3)^2=32 ?

The solutions to the equation 4(x+3)^2=32 are x=2sqrt(2)-3,x=-2sqrt(2)-3

Find the zeros of 4(x+3)^2=32

The zeros of 4(x+3)^2=32 are x=2sqrt(2)-3,x=-2sqrt(2)-3

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4 (x + 3)^{2} = 32

x = 22 - 3, x = - 22 - 3

Decimal Notation

x = - 0.17157 \dots, x = - 5.82842 \dots

Solution steps

4 (x + 3)^{2} = 32

Divide both sides by

4

(x + 3)^{2} = 8

For

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

the solutions are

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

Solve

x + 3 = 8 : x = 22 - 3

Solve

x + 3 = - 8 : x = - 22 - 3

The solutions to the quadratic equation are:

x = 22 - 3, x = - 22 - 3

t^{2} - 10 t + 9 = 0

8 x^{2} + 6 x = 0

x (x + 2) = 24

x^{2} - 8 x - 30 = 0

Q, MC = \frac{3}{2} Q^{2} - 1

What are the solutions to the equation 4(x+3)^2=32 ?
The solutions to the equation 4(x+3)^2=32 are x=2sqrt(2)-3,x=-2sqrt(2)-3
Find the zeros of 4(x+3)^2=32
The zeros of 4(x+3)^2=32 are x=2sqrt(2)-3,x=-2sqrt(2)-3