What are the solutions to the equation 12x-6x^2=4 ?

The solutions to the equation 12x-6x^2=4 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3

Find the zeros of 12x-6x^2=4

The zeros of 12x-6x^2=4 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3

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12 x - 6 x^{2} = 4

x = \frac{3 - 3}{3}, x = \frac{3 + 3}{3}

Decimal Notation

x = 0.42264 \dots, x = 1.57735 \dots

Solution steps

12 x - 6 x^{2} = 4

Move

4

to the left side

12 x - 6 x^{2} - 4 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 6 x^{2} + 12 x - 4 = 0

Solve with the quadratic formula

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

1 2^{2} - 4 (- 6) (- 4) = 43

x_{1, 2} = \frac{- 12 \pm 4 3}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- 12 + 4 3}{2 ( - 6 )}, x_{2} = \frac{- 12 - 4 3}{2 ( - 6 )}

x = \frac{- 12 + 4 3}{2 ( - 6 )} : \frac{3 - 3}{3}

x = \frac{- 12 - 4 3}{2 ( - 6 )} : \frac{3 + 3}{3}

The solutions to the quadratic equation are:

x = \frac{3 - 3}{3}, x = \frac{3 + 3}{3}

- x - 25 = - 5 x + 51

6 x^{2} - x - 1 < 0

∣ 6 x - 4 ∣ + 4 = 7

\frac{\frac{2}{5} - \frac{1}{2}}{\frac{1}{10} - \frac{3}{15}}

- 3 x + 6 x + 9

What are the solutions to the equation 12x-6x^2=4 ?
The solutions to the equation 12x-6x^2=4 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3
Find the zeros of 12x-6x^2=4
The zeros of 12x-6x^2=4 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3