What are the solutions to the equation 12x^2-128x+231=0 ?

The solutions to the equation 12x^2-128x+231=0 are x=(32+sqrt(331))/6 ,x=(32-sqrt(331))/6

Find the zeros of 12x^2-128x+231=0

The zeros of 12x^2-128x+231=0 are x=(32+sqrt(331))/6 ,x=(32-sqrt(331))/6

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12 x^{2} - 128 x + 231 = 0

x = \frac{32 + 331}{6}, x = \frac{32 - 331}{6}

Decimal Notation

x = 8.36556 \dots, x = 2.30109 \dots

Solution steps

12 x^{2} - 128 x + 231 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 128 ) \pm ( - 128 ) ^{2} - 4 \cdot 12 \cdot 231}{2 \cdot 12}

(- 128)^{2} - 4 \cdot 12 \cdot 231 = 4331

x_{1, 2} = \frac{- ( - 128 ) \pm 4 331}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 128 ) + 4 331}{2 \cdot 12}, x_{2} = \frac{- ( - 128 ) - 4 331}{2 \cdot 12}

x = \frac{- ( - 128 ) + 4 331}{2 \cdot 12} : \frac{32 + 331}{6}

x = \frac{- ( - 128 ) - 4 331}{2 \cdot 12} : \frac{32 - 331}{6}

The solutions to the quadratic equation are:

x = \frac{32 + 331}{6}, x = \frac{32 - 331}{6}

2 x^{2} - 15 x + 25 = 0

y^{2} = - 12

5 x^{2} = - 8 x

2 x = x^{2} - 8

5 r^{2} - 6 r = - 1

What are the solutions to the equation 12x^2-128x+231=0 ?
The solutions to the equation 12x^2-128x+231=0 are x=(32+sqrt(331))/6 ,x=(32-sqrt(331))/6
Find the zeros of 12x^2-128x+231=0
The zeros of 12x^2-128x+231=0 are x=(32+sqrt(331))/6 ,x=(32-sqrt(331))/6