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

The solutions to the equation 0=3x^2-12x+12 are x=2

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

The zeros of 0=3x^2-12x+12 are x=2

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

x = 2

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

(- 12)^{2} - 4 \cdot 3 \cdot 12 = 0

x_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 3}

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

\frac{- ( - 12 )}{2 \cdot 3} = 2

x = 2

The solution to the quadratic equation is:

x = 2

1 2^{2} + 1 3^{2} = c^{2}

6 x^{2} - 54 = 5 x^{2} + 171

x^{2} - 6 x + 9

x^{2} + x - 6 = 24

x^{2} + 9 + 20 = 0

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