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

The solutions to the equation 2x^2+4x-5=2(x-1) are x=(-1+sqrt(7))/2 ,x=-(1+sqrt(7))/2

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

The zeros of 2x^2+4x-5=2(x-1) are x=(-1+sqrt(7))/2 ,x=-(1+sqrt(7))/2

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2 x^{2} + 4 x - 5 = 2 (x - 1)

x = \frac{- 1 + 7}{2}, x = - \frac{1 + 7}{2}

Decimal Notation

x = 0.82287 \dots, x = - 1.82287 \dots

Solution steps

2 x^{2} + 4 x - 5 = 2 (x - 1)

Expand

2 (x - 1) : 2 x - 2

2 x^{2} + 4 x - 5 = 2 x - 2

Move

2

to the left side

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

Move

2 x

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 3) = 27

x_{1, 2} = \frac{- 2 \pm 2 7}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 7}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 7}{2 \cdot 2}

x = \frac{- 2 + 2 7}{2 \cdot 2} : \frac{- 1 + 7}{2}

x = \frac{- 2 - 2 7}{2 \cdot 2} : - \frac{1 + 7}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 7}{2}, x = - \frac{1 + 7}{2}

\frac{1}{4} (12 m + 28) = 6 + 2 (3 m + 1)

ln (e^{2})

x^{3} + 3 x

\frac{15 a - b}{12 a} - \frac{a - 4 b}{9 a}

3 x + 40 = x + 20

What are the solutions to the equation 2x^2+4x-5=2(x-1) ?
The solutions to the equation 2x^2+4x-5=2(x-1) are x=(-1+sqrt(7))/2 ,x=-(1+sqrt(7))/2
Find the zeros of 2x^2+4x-5=2(x-1)
The zeros of 2x^2+4x-5=2(x-1) are x=(-1+sqrt(7))/2 ,x=-(1+sqrt(7))/2