What are the solutions to the equation 3x^2-4x-4=2x^2-12x ?

The solutions to the equation 3x^2-4x-4=2x^2-12x are x=2(sqrt(5)-2),x=-2(2+sqrt(5))

Find the zeros of 3x^2-4x-4=2x^2-12x

The zeros of 3x^2-4x-4=2x^2-12x are x=2(sqrt(5)-2),x=-2(2+sqrt(5))

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

x = 2 (5 - 2), x = - 2 (2 + 5)

Solution steps

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

Move

12 x

to the left side

3 x^{2} + 8 x - 4 = 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} + 8 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 4) = 45

x_{1, 2} = \frac{- 8 \pm 4 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 4 5}{2 \cdot 1}, x_{2} = \frac{- 8 - 4 5}{2 \cdot 1}

x = \frac{- 8 + 4 5}{2 \cdot 1} : 2 (5 - 2)

x = \frac{- 8 - 4 5}{2 \cdot 1} : - 2 (2 + 5)

The solutions to the quadratic equation are:

x = 2 (5 - 2), x = - 2 (2 + 5)

x^{2} = \frac{16}{49}

8 x^{2} + 1 = 0

129 6^{\frac{3}{4}}

(5 x + 1)^{2} - 2 x (5 x + 1) = 0

3 x^{2} - 8 = - 155

What are the solutions to the equation 3x^2-4x-4=2x^2-12x ?
The solutions to the equation 3x^2-4x-4=2x^2-12x are x=2(sqrt(5)-2),x=-2(2+sqrt(5))
Find the zeros of 3x^2-4x-4=2x^2-12x
The zeros of 3x^2-4x-4=2x^2-12x are x=2(sqrt(5)-2),x=-2(2+sqrt(5))