What are the solutions to the equation solvefor x,2x^2-32x+128=0 ?

The solutions to the equation solvefor x,2x^2-32x+128=0 are x=8

Find the zeros of solvefor x,2x^2-32x+128=0

The zeros of solvefor x,2x^2-32x+128=0 are x=8

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x, 2 x^{2} - 32 x + 128 = 0

x = 8

Solution steps

2 x^{2} - 32 x + 128 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 2 \cdot 128}{2 \cdot 2}

(- 32)^{2} - 4 \cdot 2 \cdot 128 = 0

x_{1, 2} = \frac{- ( - 32 ) \pm 0}{2 \cdot 2}

x = \frac{- ( - 32 )}{2 \cdot 2}

\frac{- ( - 32 )}{2 \cdot 2} = 8

x = 8

The solution to the quadratic equation is:

x = 8

(2 x - 4)^{2} = 9

x^{2} - 4 = - 7

25 x^{2} - 4 = 0

z^{2} + 4 z

x^{2} + (\frac{4 - x}{2})^{2} = 16

What are the solutions to the equation solvefor x,2x^2-32x+128=0 ?
The solutions to the equation solvefor x,2x^2-32x+128=0 are x=8
Find the zeros of solvefor x,2x^2-32x+128=0
The zeros of solvefor x,2x^2-32x+128=0 are x=8