What are the solutions to the equation (2x-1)(x+2)=8x-4 ?

The solutions to the equation (2x-1)(x+2)=8x-4 are x=2,x= 1/2

Find the zeros of (2x-1)(x+2)=8x-4

The zeros of (2x-1)(x+2)=8x-4 are x=2,x= 1/2

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(2 x - 1) (x + 2) = 8 x - 4

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

Decimal

x = 2, x = 0.5

Solution steps

(2 x - 1) (x + 2) = 8 x - 4

Expand

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

2 x^{2} + 3 x - 2 = 8 x - 4

Move

4

to the left side

2 x^{2} + 3 x + 2 = 8 x

Move

8 x

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 2}

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

x = \frac{- ( - 5 ) - 3}{2 \cdot 2} : \frac{1}{2}

The solutions to the quadratic equation are:

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

3 x^{2} - 26 = 1

x^{2} - 3 x - 19 = 0

(x - 4)^{2} - 10 = (x - 1)^{2} - x^{2}

x^{2} + x + 3

x^{2} + x + 6

What are the solutions to the equation (2x-1)(x+2)=8x-4 ?
The solutions to the equation (2x-1)(x+2)=8x-4 are x=2,x= 1/2
Find the zeros of (2x-1)(x+2)=8x-4
The zeros of (2x-1)(x+2)=8x-4 are x=2,x= 1/2