What are the solutions to the equation 1/4 (2x-12)^2+8=4 ?

The solutions to the equation 1/4 (2x-12)^2+8=4 are x=6+2i,x=6-2i

Find the zeros of 1/4 (2x-12)^2+8=4

The zeros of 1/4 (2x-12)^2+8=4 are x=6+2i,x=6-2i

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\frac{1}{4} (2 x - 12)^{2} + 8 = 4

x = 6 + 2 i, x = 6 - 2 i

Solution steps

\frac{1}{4} (2 x - 12)^{2} + 8 = 4

Move

8

to the right side

\frac{1}{4} (2 x - 12)^{2} = - 4

Multiply both sides by

4

(2 x - 12)^{2} = - 16

For

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

the solutions are

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

Solve

2 x - 12 = - 16 : x = 6 + 2 i

Solve

2 x - 12 = - - 16 : x = 6 - 2 i

The solutions to the quadratic equation are:

x = 6 + 2 i, x = 6 - 2 i

5 x - x^{2}

0.32 x^{2} - 1.2 x + 1 = 0

a x^{2} + (a + b) x + b = 0

x + 3 = 39 - x^{2}

y = y (4 - y)

What are the solutions to the equation 1/4 (2x-12)^2+8=4 ?
The solutions to the equation 1/4 (2x-12)^2+8=4 are x=6+2i,x=6-2i
Find the zeros of 1/4 (2x-12)^2+8=4
The zeros of 1/4 (2x-12)^2+8=4 are x=6+2i,x=6-2i