What are the solutions to the equation 2/3 (2x-3)^2+12=8 ?

The solutions to the equation 2/3 (2x-3)^2+12=8 are x= 3/2+isqrt(3/2),x= 3/2-isqrt(3/2)

Find the zeros of 2/3 (2x-3)^2+12=8

The zeros of 2/3 (2x-3)^2+12=8 are x= 3/2+isqrt(3/2),x= 3/2-isqrt(3/2)

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\frac{2}{3} (2 x - 3)^{2} + 12 = 8

x = \frac{3}{2} + i \frac{3}{2}, x = \frac{3}{2} - i \frac{3}{2}

Solution steps

\frac{2}{3} (2 x - 3)^{2} + 12 = 8

Move

12

to the right side

\frac{2}{3} (2 x - 3)^{2} = - 4

Multiply both sides by

3

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

Divide both sides by

2

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

For

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

the solutions are

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

Solve

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

Solve

2 x - 3 = - - 6 : x = \frac{3}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{3}{2} + i \frac{3}{2}, x = \frac{3}{2} - i \frac{3}{2}

x^{2} - 30 x + 225 = 400

3 - 12 x^{2} = 0

x^{2} - 11 x + 9 = 0

- 16 t^{2} + 40 t = 0

1 6^{2} + 25 y^{2} - 400 = 0

What are the solutions to the equation 2/3 (2x-3)^2+12=8 ?
The solutions to the equation 2/3 (2x-3)^2+12=8 are x= 3/2+isqrt(3/2),x= 3/2-isqrt(3/2)
Find the zeros of 2/3 (2x-3)^2+12=8
The zeros of 2/3 (2x-3)^2+12=8 are x= 3/2+isqrt(3/2),x= 3/2-isqrt(3/2)