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

The solutions to the equation (2x-1)(2+x)=(x-2)(x+1) are x=0,x=-4

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

The zeros of (2x-1)(2+x)=(x-2)(x+1) are x=0,x=-4

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(2 x - 1) (2 + x) = (x - 2) (x + 1)

x = 0, x = - 4

Solution steps

(2 x - 1) (2 + x) = (x - 2) (x + 1)

Expand

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

Expand

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

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

Move

2

to the left side

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

Move

x

to the left side

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

Move

x^{2}

to the left side

x^{2} + 4 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot 0 = 4

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

Separate the solutions

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

x = \frac{- 4 + 4}{2 \cdot 1} : 0

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

The solutions to the quadratic equation are:

x = 0, x = - 4

x^{2} + 5 x + 2 = 4

(2 x + 3) (3 x - 7) = 0

3 (x + 4)^{2} = 8

- 4 + 16 x - 15 x^{2} = 0

4 x^{2} - 20 x + 29 = 0

What are the solutions to the equation (2x-1)(2+x)=(x-2)(x+1) ?
The solutions to the equation (2x-1)(2+x)=(x-2)(x+1) are x=0,x=-4
Find the zeros of (2x-1)(2+x)=(x-2)(x+1)
The zeros of (2x-1)(2+x)=(x-2)(x+1) are x=0,x=-4