What are the solutions to the equation 2/3 x^2-256x+128^2=0 ?

The solutions to the equation 2/3 x^2-256x+128^2=0 are x=64(3+sqrt(3)),x=64(3-sqrt(3))

Find the zeros of 2/3 x^2-256x+128^2=0

The zeros of 2/3 x^2-256x+128^2=0 are x=64(3+sqrt(3)),x=64(3-sqrt(3))

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\frac{2}{3} x^{2} - 256 x + 12 8^{2} = 0

x = 64 (3 + 3), x = 64 (3 - 3)

Decimal Notation

x = 302.85125 \dots, x = 81.14874 \dots

Solution steps

\frac{2}{3} x^{2} - 256 x + 12 8^{2} = 0

Multiply both sides by

3

2 x^{2} - 768 x + 49152 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 768 ) \pm ( - 768 ) ^{2} - 4 \cdot 2 \cdot 49152}{2 \cdot 2}

(- 768)^{2} - 4 \cdot 2 \cdot 49152 = 2563

x_{1, 2} = \frac{- ( - 768 ) \pm 256 3}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 768 ) + 256 3}{2 \cdot 2}, x_{2} = \frac{- ( - 768 ) - 256 3}{2 \cdot 2}

x = \frac{- ( - 768 ) + 256 3}{2 \cdot 2} : 64 (3 + 3)

x = \frac{- ( - 768 ) - 256 3}{2 \cdot 2} : 64 (3 - 3)

The solutions to the quadratic equation are:

x = 64 (3 + 3), x = 64 (3 - 3)

9 x^{2} + 9 x + 52 = 0

6.5 = \frac{( x + x ^{2} + 5 x )}{2}

2 t^{2} - 2 t + 1

b, \frac{4 x ^{2} + b ^{2} - 2 b x}{3} = y

0 = a^{2} - 2 a + 2.34

What are the solutions to the equation 2/3 x^2-256x+128^2=0 ?
The solutions to the equation 2/3 x^2-256x+128^2=0 are x=64(3+sqrt(3)),x=64(3-sqrt(3))
Find the zeros of 2/3 x^2-256x+128^2=0
The zeros of 2/3 x^2-256x+128^2=0 are x=64(3+sqrt(3)),x=64(3-sqrt(3))