What are the solutions to the equation 3(x-3)^2+12=0 ?

The solutions to the equation 3(x-3)^2+12=0 are x=2i+3,x=-2i+3

Find the zeros of 3(x-3)^2+12=0

The zeros of 3(x-3)^2+12=0 are x=2i+3,x=-2i+3

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3 (x - 3)^{2} + 12 = 0

x = 2 i + 3, x = - 2 i + 3

Solution steps

3 (x - 3)^{2} + 12 = 0

Move

12

to the right side

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

Divide both sides by

3

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

For

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

the solutions are

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

Solve

x - 3 = - 4 : x = 2 i + 3

Solve

x - 3 = - - 4 : x = - 2 i + 3

The solutions to the quadratic equation are:

x = 2 i + 3, x = - 2 i + 3

8 x^{2} - 28 x - 16

2 x \cdot 4 x^{- \frac{5}{2}}

\frac{5 x ^{2}}{6} = \frac{6}{125}

((x^{2} - 1)^{\frac{1}{2}})^{2}

What are the solutions to the equation 3(x-3)^2+12=0 ?
The solutions to the equation 3(x-3)^2+12=0 are x=2i+3,x=-2i+3
Find the zeros of 3(x-3)^2+12=0
The zeros of 3(x-3)^2+12=0 are x=2i+3,x=-2i+3