What are the solutions to the equation -12x^2+48x-16=0 ?

The solutions to the equation -12x^2+48x-16=0 are x=(2(3-sqrt(6)))/3 ,x=(2(3+sqrt(6)))/3

Find the zeros of -12x^2+48x-16=0

The zeros of -12x^2+48x-16=0 are x=(2(3-sqrt(6)))/3 ,x=(2(3+sqrt(6)))/3

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- 12 x^{2} + 48 x - 16 = 0

x = \frac{2 ( 3 - 6 )}{3}, x = \frac{2 ( 3 + 6 )}{3}

Decimal Notation

x = 0.36700 \dots, x = 3.63299 \dots

Solution steps

- 12 x^{2} + 48 x - 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 12 ) ( - 16 )}{2 ( - 12 )}

4 8^{2} - 4 (- 12) (- 16) = 166

x_{1, 2} = \frac{- 48 \pm 16 6}{2 ( - 12 )}

Separate the solutions

x_{1} = \frac{- 48 + 16 6}{2 ( - 12 )}, x_{2} = \frac{- 48 - 16 6}{2 ( - 12 )}

x = \frac{- 48 + 16 6}{2 ( - 12 )} : \frac{2 ( 3 - 6 )}{3}

x = \frac{- 48 - 16 6}{2 ( - 12 )} : \frac{2 ( 3 + 6 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 3 - 6 )}{3}, x = \frac{2 ( 3 + 6 )}{3}

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

2 x^{2} + 19 x + 45 = 0

t^{2} - 2 t - 24 = 0

x^{2} - (k - 3) x + k = 0

81 - y^{2} = 0

What are the solutions to the equation -12x^2+48x-16=0 ?
The solutions to the equation -12x^2+48x-16=0 are x=(2(3-sqrt(6)))/3 ,x=(2(3+sqrt(6)))/3
Find the zeros of -12x^2+48x-16=0
The zeros of -12x^2+48x-16=0 are x=(2(3-sqrt(6)))/3 ,x=(2(3+sqrt(6)))/3