What are the solutions to the equation 2x^2-4x=12x-32 ?

The solutions to the equation 2x^2-4x=12x-32 are x=4

Find the zeros of 2x^2-4x=12x-32

The zeros of 2x^2-4x=12x-32 are x=4

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2 x^{2} - 4 x = 12 x - 32

x = 4

Solution steps

2 x^{2} - 4 x = 12 x - 32

Move

32

to the left side

2 x^{2} - 4 x + 32 = 12 x

Move

12 x

to the left side

2 x^{2} - 16 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 2 \cdot 32}{2 \cdot 2}

(- 16)^{2} - 4 \cdot 2 \cdot 32 = 0

x_{1, 2} = \frac{- ( - 16 ) \pm 0}{2 \cdot 2}

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

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

x = 4

The solution to the quadratic equation is:

x = 4

(x + 1)^{2} = 2 (x - 3)

x + 3 = (2 - x)^{2}

x^{2} - 4 x + 3 x - 2 = 28

6 a^{2} + 7 a - 3 = 0

12 x^{2} + 40 x - 32 = 0

What are the solutions to the equation 2x^2-4x=12x-32 ?
The solutions to the equation 2x^2-4x=12x-32 are x=4
Find the zeros of 2x^2-4x=12x-32
The zeros of 2x^2-4x=12x-32 are x=4