What are the solutions to the equation 12x^2-128x+240=0 ?

The solutions to the equation 12x^2-128x+240=0 are x=(2(8+sqrt(19)))/3 ,x=(2(8-sqrt(19)))/3

Find the zeros of 12x^2-128x+240=0

The zeros of 12x^2-128x+240=0 are x=(2(8+sqrt(19)))/3 ,x=(2(8-sqrt(19)))/3

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12 x^{2} - 128 x + 240 = 0

x = \frac{2 ( 8 + 19 )}{3}, x = \frac{2 ( 8 - 19 )}{3}

Decimal Notation

x = 8.23926 \dots, x = 2.42740 \dots

Solution steps

12 x^{2} - 128 x + 240 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 128 ) \pm ( - 128 ) ^{2} - 4 \cdot 12 \cdot 240}{2 \cdot 12}

(- 128)^{2} - 4 \cdot 12 \cdot 240 = 1619

x_{1, 2} = \frac{- ( - 128 ) \pm 16 19}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- ( - 128 ) + 16 19}{2 \cdot 12}, x_{2} = \frac{- ( - 128 ) - 16 19}{2 \cdot 12}

x = \frac{- ( - 128 ) + 16 19}{2 \cdot 12} : \frac{2 ( 8 + 19 )}{3}

x = \frac{- ( - 128 ) - 16 19}{2 \cdot 12} : \frac{2 ( 8 - 19 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 8 + 19 )}{3}, x = \frac{2 ( 8 - 19 )}{3}

5 x^{2} + 12 x + 4 = 0

2 x^{2} - 20 = 10

x^{2} + 4 x - 19 = 0

3 x^{2} = 9 x

2 x^{2} - 24 = 26

What are the solutions to the equation 12x^2-128x+240=0 ?
The solutions to the equation 12x^2-128x+240=0 are x=(2(8+sqrt(19)))/3 ,x=(2(8-sqrt(19)))/3
Find the zeros of 12x^2-128x+240=0
The zeros of 12x^2-128x+240=0 are x=(2(8+sqrt(19)))/3 ,x=(2(8-sqrt(19)))/3