What are the solutions to the equation 12x^2-48x+32=0 ?

The solutions to the equation 12x^2-48x+32=0 are x=(2(3+sqrt(3)))/3 ,x=(2(3-sqrt(3)))/3

Find the zeros of 12x^2-48x+32=0

The zeros of 12x^2-48x+32=0 are x=(2(3+sqrt(3)))/3 ,x=(2(3-sqrt(3)))/3

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12 x^{2} - 48 x + 32 = 0

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

Decimal Notation

x = 3.15470 \dots, x = 0.84529 \dots

Solution steps

12 x^{2} - 48 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 12 \cdot 32}{2 \cdot 12}

(- 48)^{2} - 4 \cdot 12 \cdot 32 = 163

x_{1, 2} = \frac{- ( - 48 ) \pm 16 3}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 16 3}{2 \cdot 12}, x_{2} = \frac{- ( - 48 ) - 16 3}{2 \cdot 12}

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

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

The solutions to the quadratic equation are:

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

2 w^{2} + 7 w - 3 = 0

20 x^{2} - 9 x + 1 = 0

1 + 2 b = x^{2} + b x + b

w^{2} + 2 w - 8 = 0

0 = - 16 t^{2} + 4624

What are the solutions to the equation 12x^2-48x+32=0 ?
The solutions to the equation 12x^2-48x+32=0 are x=(2(3+sqrt(3)))/3 ,x=(2(3-sqrt(3)))/3
Find the zeros of 12x^2-48x+32=0
The zeros of 12x^2-48x+32=0 are x=(2(3+sqrt(3)))/3 ,x=(2(3-sqrt(3)))/3