What are the solutions to the equation (x-1)(x+1)=2x+(5x^2-3) ?

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

Find the zeros of (x-1)(x+1)=2x+(5x^2-3)

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

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

x = \frac{1}{2}, x = - 1

Solution steps

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

Expand

(x - 1) (x + 1) : x^{2} - 1

x^{2} - 1 = 2 x + 5 x^{2} - 3

Switch sides

2 x + 5 x^{2} - 3 = x^{2} - 1

Move

1

to the left side

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

Move

x^{2}

to the left side

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

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 2 + 6}{2 \cdot 4}, x_{2} = \frac{- 2 - 6}{2 \cdot 4}

x = \frac{- 2 + 6}{2 \cdot 4} : \frac{1}{2}

x = \frac{- 2 - 6}{2 \cdot 4} : - 1

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - 1

2 x^{2} - x + 10 = 0

96 = 2 x + (2 x + 12)

2^{\frac{4}{2}}

∣ 5 x ∣ \leq 10

6 y + 2 = 9 y + 14

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