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

The solutions to the equation 3x^2-12x+12=3 are x=3,x=1

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

The zeros of 3x^2-12x+12=3 are x=3,x=1

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

x = 3, x = 1

Solution steps

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

Move

3

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 9}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 9 = 6

x_{1, 2} = \frac{- ( - 12 ) \pm 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 6}{2 \cdot 3}, x_{2} = \frac{- ( - 12 ) - 6}{2 \cdot 3}

x = \frac{- ( - 12 ) + 6}{2 \cdot 3} : 3

x = \frac{- ( - 12 ) - 6}{2 \cdot 3} : 1

The solutions to the quadratic equation are:

x = 3, x = 1

2 + - 25

d^{2} + 56 = - 18 d

3 t^{2} - 12 = 0

2 x - 5 x (2 - x) = 3 x - 7

23.8 = 3.5 b

What are the solutions to the equation 3x^2-12x+12=3 ?
The solutions to the equation 3x^2-12x+12=3 are x=3,x=1
Find the zeros of 3x^2-12x+12=3
The zeros of 3x^2-12x+12=3 are x=3,x=1