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

The solutions to the equation 2x(x-1)=6 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2

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

The zeros of 2x(x-1)=6 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2

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2 x (x - 1) = 6

x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

Decimal Notation

x = 2.30277 \dots, x = - 1.30277 \dots

Solution steps

2 x (x - 1) = 6

Expand

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

2 x^{2} - 2 x = 6

Move

6

to the left side

2 x^{2} - 2 x - 6 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 6) = 213

x_{1, 2} = \frac{- ( - 2 ) \pm 2 13}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 13}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 13}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 13}{2 \cdot 2} : \frac{1 + 13}{2}

x = \frac{- ( - 2 ) - 2 13}{2 \cdot 2} : \frac{1 - 13}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 13}{2}, x = \frac{1 - 13}{2}

x^{2} + 4 = 18

0.006 q^{2} + 50 = 60

36 - x^{2} = 6 + \frac{x ^{2}}{4}

3 x^{2} + 4 x = 10

3 x^{2} + 2 x = - 3 x + 5

What are the solutions to the equation 2x(x-1)=6 ?
The solutions to the equation 2x(x-1)=6 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2
Find the zeros of 2x(x-1)=6
The zeros of 2x(x-1)=6 are x=(1+sqrt(13))/2 ,x=(1-sqrt(13))/2