What are the solutions to the equation 9(2x+8)^2+41=42 ?

The solutions to the equation 9(2x+8)^2+41=42 are x=-23/6 ,x=-25/6

Find the zeros of 9(2x+8)^2+41=42

The zeros of 9(2x+8)^2+41=42 are x=-23/6 ,x=-25/6

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9 (2 x + 8)^{2} + 41 = 42

x = - \frac{23}{6}, x = - \frac{25}{6}

Decimal Notation

x = - 3.83333 \dots, x = - 4.16666 \dots

Solution steps

9 (2 x + 8)^{2} + 41 = 42

Move

41

to the right side

9 (2 x + 8)^{2} = 1

Divide both sides by

9

(2 x + 8)^{2} = \frac{1}{9}

For

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

the solutions are

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

Solve

2 x + 8 = \frac{1}{9} : x = - \frac{23}{6}

Solve

2 x + 8 = - \frac{1}{9} : x = - \frac{25}{6}

The solutions to the quadratic equation are:

x = - \frac{23}{6}, x = - \frac{25}{6}

- 2 x^{2} - 4 x + 1 = 0

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

7 x^{2} - 105 = 0

2 (3 x^{2} - 1) = - x

- 2 n^{2} = 14

What are the solutions to the equation 9(2x+8)^2+41=42 ?
The solutions to the equation 9(2x+8)^2+41=42 are x=-23/6 ,x=-25/6
Find the zeros of 9(2x+8)^2+41=42
The zeros of 9(2x+8)^2+41=42 are x=-23/6 ,x=-25/6