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

The solutions to the equation (x-1)(2x-2)=3 are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2

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

The zeros of (x-1)(2x-2)=3 are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2

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

x = \frac{2 + 6}{2}, x = \frac{2 - 6}{2}

Decimal Notation

x = 2.22474 \dots, x = - 0.22474 \dots

Solution steps

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

Expand

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

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

Move

3

to the left side

2 x^{2} - 4 x - 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 (- 1) = 26

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{2 + 6}{2}, x = \frac{2 - 6}{2}

8 t (t + 3) = 2 t - 5

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

x, x^{2} = 2 y^{2} ln (y)

3 x^{2} + 1 = 121

x^{2} + 8 x - 84 = 0

What are the solutions to the equation (x-1)(2x-2)=3 ?
The solutions to the equation (x-1)(2x-2)=3 are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2
Find the zeros of (x-1)(2x-2)=3
The zeros of (x-1)(2x-2)=3 are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2