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

The solutions to the equation 3x^2-12x+12=(x-3) are x= 13/6+i(sqrt(11))/6 ,x= 13/6-i(sqrt(11))/6

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

The zeros of 3x^2-12x+12=(x-3) are x= 13/6+i(sqrt(11))/6 ,x= 13/6-i(sqrt(11))/6

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

x = \frac{13}{6} + i \frac{11}{6}, x = \frac{13}{6} - i \frac{11}{6}

Solution steps

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

Refine

3 x^{2} - 12 x + 12 = x - 3

Move

3

to the left side

3 x^{2} - 12 x + 15 = x

Move

x

to the left side

3 x^{2} - 13 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 3 \cdot 15}{2 \cdot 3}

Simplify

(- 13)^{2} - 4 \cdot 3 \cdot 15 : 11 i

x_{1, 2} = \frac{- ( - 13 ) \pm 11 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 11 i}{2 \cdot 3}, x_{2} = \frac{- ( - 13 ) - 11 i}{2 \cdot 3}

x = \frac{- ( - 13 ) + 11 i}{2 \cdot 3} : \frac{13}{6} + i \frac{11}{6}

x = \frac{- ( - 13 ) - 11 i}{2 \cdot 3} : \frac{13}{6} - i \frac{11}{6}

The solutions to the quadratic equation are:

x = \frac{13}{6} + i \frac{11}{6}, x = \frac{13}{6} - i \frac{11}{6}

12 x^{2} - 6 x + 9 = 0

0 = 5 x^{2} - 16

s^{2} + 4 s + 7 = 0

2 x^{2} - 5 x = - 1

2 x (x - \frac{7}{2}) = 49

What are the solutions to the equation 3x^2-12x+12=(x-3) ?
The solutions to the equation 3x^2-12x+12=(x-3) are x= 13/6+i(sqrt(11))/6 ,x= 13/6-i(sqrt(11))/6
Find the zeros of 3x^2-12x+12=(x-3)
The zeros of 3x^2-12x+12=(x-3) are x= 13/6+i(sqrt(11))/6 ,x= 13/6-i(sqrt(11))/6