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

The solutions to the equation (x-2)(x+1)=3x-2 are x=4,x=0

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

The zeros of (x-2)(x+1)=3x-2 are x=4,x=0

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(x - 2) (x + 1) = 3 x - 2

x = 4, x = 0

Solution steps

(x - 2) (x + 1) = 3 x - 2

Expand

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

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

Move

2

to the left side

x^{2} - x = 3 x

Move

3 x

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} : 4

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

The solutions to the quadratic equation are:

x = 4, x = 0

36 (\frac{1}{3})

(6 x + 2) (x - 1) \leq (2 x - 3) (3 x - 2)

3 (1 - 9 a) + 22 a = 2 (2 a - 9) - 15

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

25 x^{2} + 70 x + 49

What are the solutions to the equation (x-2)(x+1)=3x-2 ?
The solutions to the equation (x-2)(x+1)=3x-2 are x=4,x=0
Find the zeros of (x-2)(x+1)=3x-2
The zeros of (x-2)(x+1)=3x-2 are x=4,x=0