What are the solutions to the equation 480=(6x-4)(4x-4)(2) ?

The solutions to the equation 480=(6x-4)(4x-4)(2) are x=4,x=-7/3

Find the zeros of 480=(6x-4)(4x-4)(2)

The zeros of 480=(6x-4)(4x-4)(2) are x=4,x=-7/3

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480 = (6 x - 4) (4 x - 4) (2)

x = 4, x = - \frac{7}{3}

Decimal Notation

x = 4, x = - 2.33333 \dots

Solution steps

480 = (6 x - 4) (4 x - 4) (2)

Expand

(6 x - 4) (4 x - 4) (2) : 48 x^{2} - 80 x + 32

480 = 48 x^{2} - 80 x + 32

Switch sides

48 x^{2} - 80 x + 32 = 480

Move

480

to the left side

48 x^{2} - 80 x - 448 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 80 ) \pm ( - 80 ) ^{2} - 4 \cdot 48 ( - 448 )}{2 \cdot 48}

(- 80)^{2} - 4 \cdot 48 (- 448) = 304

x_{1, 2} = \frac{- ( - 80 ) \pm 304}{2 \cdot 48}

Separate the solutions

x_{1} = \frac{- ( - 80 ) + 304}{2 \cdot 48}, x_{2} = \frac{- ( - 80 ) - 304}{2 \cdot 48}

x = \frac{- ( - 80 ) + 304}{2 \cdot 48} : 4

x = \frac{- ( - 80 ) - 304}{2 \cdot 48} : - \frac{7}{3}

The solutions to the quadratic equation are:

x = 4, x = - \frac{7}{3}

x^{2} + (- 282) = 51

x (x + 1) - 2 (x + 1) = 0

D^{2} + 9 = 0

x^{2} - 32 = 4 x

x^{2} + 16 x + 0 = 0

What are the solutions to the equation 480=(6x-4)(4x-4)(2) ?
The solutions to the equation 480=(6x-4)(4x-4)(2) are x=4,x=-7/3
Find the zeros of 480=(6x-4)(4x-4)(2)
The zeros of 480=(6x-4)(4x-4)(2) are x=4,x=-7/3