What are the solutions to the equation 2x-4=(x+1)(2x-4)-5(x+1) ?

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

Find the zeros of 2x-4=(x+1)(2x-4)-5(x+1)

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

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2 x - 4 = (x + 1) (2 x - 4) - 5 (x + 1)

x = 5, x = - \frac{1}{2}

Decimal Notation

x = 5, x = - 0.5

Solution steps

2 x - 4 = (x + 1) (2 x - 4) - 5 (x + 1)

Expand

(x + 1) (2 x - 4) - 5 (x + 1) : 2 x^{2} - 7 x - 9

2 x - 4 = 2 x^{2} - 7 x - 9

Switch sides

2 x^{2} - 7 x - 9 = 2 x - 4

Move

4

to the left side

2 x^{2} - 7 x - 5 = 2 x

Move

2 x

to the left side

2 x^{2} - 9 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 (- 5) = 11

x_{1, 2} = \frac{- ( - 9 ) \pm 11}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 11}{2 \cdot 2}, x_{2} = \frac{- ( - 9 ) - 11}{2 \cdot 2}

x = \frac{- ( - 9 ) + 11}{2 \cdot 2} : 5

x = \frac{- ( - 9 ) - 11}{2 \cdot 2} : - \frac{1}{2}

The solutions to the quadratic equation are:

x = 5, x = - \frac{1}{2}

10 y^{2} - 25 y = 0

x^{2} = 7^{2} + 8^{2}

0 = 3 x^{2} - 8 x - 3

\frac{1}{2} x^{2} - 2 x - 6 = 0

y + 1 = \frac{y ^{2} - 6}{2}

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