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

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

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

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

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

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

Write in the standard form

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

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}

x, x^{2} + y^{2} - 6 x + 4 y + 14 = 0

4 x^{2} - 2 x + 7 = 0

v^{2} - 5 v - 24 = 0

x, 5 y^{3} = x^{2}

y^{2} - y - 56 = 0

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