What are the solutions to the equation -2(6x-3)^2-12=-48 ?

The solutions to the equation -2(6x-3)^2-12=-48 are x=(sqrt(2)+1)/2 ,x=(-sqrt(2)+1)/2

Find the zeros of -2(6x-3)^2-12=-48

The zeros of -2(6x-3)^2-12=-48 are x=(sqrt(2)+1)/2 ,x=(-sqrt(2)+1)/2

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- 2 (6 x - 3)^{2} - 12 = - 48

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

Decimal Notation

x = 1.20710 \dots, x = - 0.20710 \dots

Solution steps

- 2 (6 x - 3)^{2} - 12 = - 48

Move

12

to the right side

- 2 (6 x - 3)^{2} = - 36

Divide both sides by

- 2

(6 x - 3)^{2} = 18

For

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

the solutions are

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

Solve

6 x - 3 = 18 : x = \frac{2 + 1}{2}

Solve

6 x - 3 = - 18 : x = \frac{- 2 + 1}{2}

The solutions to the quadratic equation are:

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

42 z^{2} - zy - y^{2}

2 x^{2} = 5

x^{2} - 3 x + 5 = 0

3 x - 7 = 17

\frac{2}{77} \times \frac{35}{3}

What are the solutions to the equation -2(6x-3)^2-12=-48 ?
The solutions to the equation -2(6x-3)^2-12=-48 are x=(sqrt(2)+1)/2 ,x=(-sqrt(2)+1)/2
Find the zeros of -2(6x-3)^2-12=-48
The zeros of -2(6x-3)^2-12=-48 are x=(sqrt(2)+1)/2 ,x=(-sqrt(2)+1)/2