What are the solutions to the equation (x+1)(2x-4)=6x ?

The solutions to the equation (x+1)(2x-4)=6x are x=2+sqrt(6),x=2-sqrt(6)

Find the zeros of (x+1)(2x-4)=6x

The zeros of (x+1)(2x-4)=6x are x=2+sqrt(6),x=2-sqrt(6)

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(x + 1) (2 x - 4) = 6 x

x = 2 + 6, x = 2 - 6

Decimal Notation

x = 4.44948 \dots, x = - 0.44948 \dots

Solution steps

(x + 1) (2 x - 4) = 6 x

Expand

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

2 x^{2} - 2 x - 4 = 6 x

Move

6 x

to the left side

2 x^{2} - 8 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 (- 4) = 46

x_{1, 2} = \frac{- ( - 8 ) \pm 4 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4 6}{2 \cdot 2}, x_{2} = \frac{- ( - 8 ) - 4 6}{2 \cdot 2}

x = \frac{- ( - 8 ) + 4 6}{2 \cdot 2} : 2 + 6

x = \frac{- ( - 8 ) - 4 6}{2 \cdot 2} : 2 - 6

The solutions to the quadratic equation are:

x = 2 + 6, x = 2 - 6

x^{2} + 8 x + 23 = 0

r^{2} + 10 r - 45 = - 6

x, x^{2} + y^{2} - 6 x + 14 y + 62 = 0

x^{2} = - 10 x

(3 x - 4) (x - 2) - 5 = 0

What are the solutions to the equation (x+1)(2x-4)=6x ?
The solutions to the equation (x+1)(2x-4)=6x are x=2+sqrt(6),x=2-sqrt(6)
Find the zeros of (x+1)(2x-4)=6x
The zeros of (x+1)(2x-4)=6x are x=2+sqrt(6),x=2-sqrt(6)