What are the solutions to the equation 16(1+3x)^2+11=36 ?

The solutions to the equation 16(1+3x)^2+11=36 are x= 1/12 ,x=-3/4

Find the zeros of 16(1+3x)^2+11=36

The zeros of 16(1+3x)^2+11=36 are x= 1/12 ,x=-3/4

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16 (1 + 3 x)^{2} + 11 = 36

x = \frac{1}{12}, x = - \frac{3}{4}

Solution steps

16 (1 + 3 x)^{2} + 11 = 36

Move

11

to the right side

16 (1 + 3 x)^{2} = 25

Divide both sides by

16

(1 + 3 x)^{2} = \frac{25}{16}

For

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

the solutions are

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

Solve

1 + 3 x = \frac{25}{16} : x = \frac{1}{12}

Solve

1 + 3 x = - \frac{25}{16} : x = - \frac{3}{4}

The solutions to the quadratic equation are:

x = \frac{1}{12}, x = - \frac{3}{4}

(x - 3) (x + 3) = 0

2 x^{2} + 8 x + 16 = 0

x^{2} + 2 x - 1680 = 0

3 x^{2} - 6 x - 15 = 12 x + 12

x, x^{2} y + 4 y - 8 = 0

What are the solutions to the equation 16(1+3x)^2+11=36 ?
The solutions to the equation 16(1+3x)^2+11=36 are x= 1/12 ,x=-3/4
Find the zeros of 16(1+3x)^2+11=36
The zeros of 16(1+3x)^2+11=36 are x= 1/12 ,x=-3/4