What are the solutions to the equation 4(2x+2)^2=36 ?

The solutions to the equation 4(2x+2)^2=36 are x= 1/2 ,x=-5/2

Find the zeros of 4(2x+2)^2=36

The zeros of 4(2x+2)^2=36 are x= 1/2 ,x=-5/2

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4 (2 x + 2)^{2} = 36

x = \frac{1}{2}, x = - \frac{5}{2}

Solution steps

4 (2 x + 2)^{2} = 36

Divide both sides by

4

(2 x + 2)^{2} = 9

For

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

the solutions are

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

Solve

2 x + 2 = 9 : x = \frac{1}{2}

Solve

2 x + 2 = - 9 : x = - \frac{5}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - \frac{5}{2}

v^{2} + 6 v - 59 = 0

x, 9 x^{2} - 4 x y + 6 y^{2} - 10 x - 20 y - 5 = 0

3 x^{2} - 21 x + 14 = 0

7 x^{2} + 13 x = 5

16 x^{2} - 8 x - 3 = 0

What are the solutions to the equation 4(2x+2)^2=36 ?
The solutions to the equation 4(2x+2)^2=36 are x= 1/2 ,x=-5/2
Find the zeros of 4(2x+2)^2=36
The zeros of 4(2x+2)^2=36 are x= 1/2 ,x=-5/2