What are the solutions to the equation 4x^2-48x+128=0 ?

The solutions to the equation 4x^2-48x+128=0 are x=8,x=4

Find the zeros of 4x^2-48x+128=0

The zeros of 4x^2-48x+128=0 are x=8,x=4

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4 x^{2} - 48 x + 128 = 0

x = 8, x = 4

Solution steps

4 x^{2} - 48 x + 128 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 ) \pm ( - 48 ) ^{2} - 4 \cdot 4 \cdot 128}{2 \cdot 4}

(- 48)^{2} - 4 \cdot 4 \cdot 128 = 16

x_{1, 2} = \frac{- ( - 48 ) \pm 16}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 48 ) + 16}{2 \cdot 4}, x_{2} = \frac{- ( - 48 ) - 16}{2 \cdot 4}

x = \frac{- ( - 48 ) + 16}{2 \cdot 4} : 8

x = \frac{- ( - 48 ) - 16}{2 \cdot 4} : 4

The solutions to the quadratic equation are:

x = 8, x = 4

x^{2} + 2 x - 99 = 0

t^{2} - 2 t + 8 = 0

- 31 = 2 - 3 a^{2}

x, 3 x^{2} - 2 x + 2 = 0

x, f^{- 1} (x) = x^{2} + 5

What are the solutions to the equation 4x^2-48x+128=0 ?
The solutions to the equation 4x^2-48x+128=0 are x=8,x=4
Find the zeros of 4x^2-48x+128=0
The zeros of 4x^2-48x+128=0 are x=8,x=4