What are the solutions to the equation -2x=(x+1)(x-1)+(4x^2-2) ?

The solutions to the equation -2x=(x+1)(x-1)+(4x^2-2) are x= 3/5 ,x=-1

Find the zeros of -2x=(x+1)(x-1)+(4x^2-2)

The zeros of -2x=(x+1)(x-1)+(4x^2-2) are x= 3/5 ,x=-1

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- 2 x = (x + 1) (x - 1) + (4 x^{2} - 2)

x = \frac{3}{5}, x = - 1

Decimal Notation

x = 0.6, x = - 1

Solution steps

- 2 x = (x + 1) (x - 1) + (4 x^{2} - 2)

Expand

(x + 1) (x - 1) + (4 x^{2} - 2) : 5 x^{2} - 3

- 2 x = 5 x^{2} - 3

Switch sides

5 x^{2} - 3 = - 2 x

Move

2 x

to the left side

5 x^{2} - 3 + 2 x = 0

Write in the standard form

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

5 x^{2} + 2 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

2^{2} - 4 \cdot 5 (- 3) = 8

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

Separate the solutions

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

x = \frac{- 2 + 8}{2 \cdot 5} : \frac{3}{5}

x = \frac{- 2 - 8}{2 \cdot 5} : - 1

The solutions to the quadratic equation are:

x = \frac{3}{5}, x = - 1

\frac{1}{4} + \frac{1}{7}

\frac{1}{\frac{2}{4}}

(3.2 \cdot 1 0^{- 5}) (2 - x)^{2} = (6.9 \cdot 1 0^{- 6}) x^{2}

3 x - 2 = 10

What are the solutions to the equation -2x=(x+1)(x-1)+(4x^2-2) ?
The solutions to the equation -2x=(x+1)(x-1)+(4x^2-2) are x= 3/5 ,x=-1
Find the zeros of -2x=(x+1)(x-1)+(4x^2-2)
The zeros of -2x=(x+1)(x-1)+(4x^2-2) are x= 3/5 ,x=-1