What are the solutions to the equation 0=3x^2-12x+9 ?

The solutions to the equation 0=3x^2-12x+9 are x=3,x=1

Find the zeros of 0=3x^2-12x+9

The zeros of 0=3x^2-12x+9 are x=3,x=1

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

x = 3, x = 1

Solution steps

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

Switch sides

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

625 - 16 t^{2}

(1.6332 - 2)^{2} + (3.667 - 1)^{2} + (1 - 1.5)^{2} \leq 4

5 x^{2} + 18 x \leq 8

8 (\frac{x}{4} - 6) \geq 4

15 x^{2} + 8 x - 12

What are the solutions to the equation 0=3x^2-12x+9 ?
The solutions to the equation 0=3x^2-12x+9 are x=3,x=1
Find the zeros of 0=3x^2-12x+9
The zeros of 0=3x^2-12x+9 are x=3,x=1