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

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

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

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

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

x = 2, x = 1

Solution steps

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

x (x - 4) = 21

2 x^{2} - 9 x + 8 = 0

7 a^{2} - 28 b^{2}

6 + 23 + 27 - 12

(x + 5)^{2} \leq 1 0^{2}

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