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

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

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

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

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

x = 1, x = 2

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 6 ) ( - 12 )}{2 ( - 6 )}

1 8^{2} - 4 (- 6) (- 12) = 6

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1, x = 2

(x - 2)^{2} - 10 = 26

(x^{2} - 4 x + 1) = 0

6 x^{2} = 8 x

x, z = - 3 x^{2} + y^{2} + 6 x - 8 y

x, s x (x) - x (0^{-}) 3 x (s) = 4 \frac{1}{( s ^{1} )}

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