What are the solutions to the equation 0=6x^2-18x+12 ?

The solutions to the equation 0=6x^2-18x+12 are x=2,x=1

Find the zeros of 0=6x^2-18x+12

The zeros of 0=6x^2-18x+12 are x=2,x=1

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0 = 6 x^{2} - 18 x + 12

x = 2, x = 1

Solution steps

0 = 6 x^{2} - 18 x + 12

Switch sides

6 x^{2} - 18 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 6 \cdot 12}{2 \cdot 6}

(- 18)^{2} - 4 \cdot 6 \cdot 12 = 6

x_{1, 2} = \frac{- ( - 18 ) \pm 6}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 6}{2 \cdot 6}, x_{2} = \frac{- ( - 18 ) - 6}{2 \cdot 6}

x = \frac{- ( - 18 ) + 6}{2 \cdot 6} : 2

x = \frac{- ( - 18 ) - 6}{2 \cdot 6} : 1

The solutions to the quadratic equation are:

x = 2, x = 1

20 x^{2} - 4 x = 0

x^{2} - 10 x + 25 = 54

15 x^{2} - 27 x - 6 = 0

3 x^{2} - 13 x = 0

5 x (x - 3) - x + 3 = 0

What are the solutions to the equation 0=6x^2-18x+12 ?
The solutions to the equation 0=6x^2-18x+12 are x=2,x=1
Find the zeros of 0=6x^2-18x+12
The zeros of 0=6x^2-18x+12 are x=2,x=1