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

The solutions to the equation 2x^2=12x-4x^2 are x=0,x=2

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

The zeros of 2x^2=12x-4x^2 are x=0,x=2

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

x = 0, x = 2

Solution steps

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

Switch sides

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

Move

2 x^{2}

to the left side

12 x - 6 x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 6 x^{2} + 12 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 6 ) \cdot 0}{2 ( - 6 )}

1 2^{2} - 4 (- 6) \cdot 0 = 12

x_{1, 2} = \frac{- 12 \pm 12}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- 12 + 12}{2 ( - 6 )}, x_{2} = \frac{- 12 - 12}{2 ( - 6 )}

x = \frac{- 12 + 12}{2 ( - 6 )} : 0

x = \frac{- 12 - 12}{2 ( - 6 )} : 2

The solutions to the quadratic equation are:

x = 0, x = 2

- \frac{1}{4} + \frac{1}{2}

\frac{7}{8} x - 10 < 14

\frac{1}{2} \cdot 3 \cdot 6

3 x + 4 = 19

- 4 x^{2} = - 36 x + 80

What are the solutions to the equation 2x^2=12x-4x^2 ?
The solutions to the equation 2x^2=12x-4x^2 are x=0,x=2
Find the zeros of 2x^2=12x-4x^2
The zeros of 2x^2=12x-4x^2 are x=0,x=2