What are the solutions to the equation 2x^2+14x-4=-x^2+3x ?

The solutions to the equation 2x^2+14x-4=-x^2+3x are x= 1/3 ,x=-4

Find the zeros of 2x^2+14x-4=-x^2+3x

The zeros of 2x^2+14x-4=-x^2+3x are x= 1/3 ,x=-4

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2 x^{2} + 14 x - 4 = - x^{2} + 3 x

x = \frac{1}{3}, x = - 4

Decimal Notation

x = 0.33333 \dots, x = - 4

Solution steps

2 x^{2} + 14 x - 4 = - x^{2} + 3 x

Move

3 x

to the left side

2 x^{2} + 11 x - 4 = - x^{2}

Move

x^{2}

to the left side

3 x^{2} + 11 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 3 ( - 4 )}{2 \cdot 3}

1 1^{2} - 4 \cdot 3 (- 4) = 13

x_{1, 2} = \frac{- 11 \pm 13}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 11 + 13}{2 \cdot 3}, x_{2} = \frac{- 11 - 13}{2 \cdot 3}

x = \frac{- 11 + 13}{2 \cdot 3} : \frac{1}{3}

x = \frac{- 11 - 13}{2 \cdot 3} : - 4

The solutions to the quadratic equation are:

x = \frac{1}{3}, x = - 4

x^{2} + 12 x - 27 = 6 x

x, 2 x^{2} - 5 x y + 2 y^{2} - 11 x - 7 y - 4 = 0

(- \frac{4}{3} b + \frac{9}{2})^{2} + b^{2} - (- \frac{4}{3} b + \frac{9}{2}) - 6 b = - 3

2 y^{2} + 11 y + 12 = 0

x^{2} + 49 = - 14 x

What are the solutions to the equation 2x^2+14x-4=-x^2+3x ?
The solutions to the equation 2x^2+14x-4=-x^2+3x are x= 1/3 ,x=-4
Find the zeros of 2x^2+14x-4=-x^2+3x
The zeros of 2x^2+14x-4=-x^2+3x are x= 1/3 ,x=-4