What are the solutions to the equation 2x^2+4x-9=2(x-1) ?

The solutions to the equation 2x^2+4x-9=2(x-1) are x=(-1+sqrt(15))/2 ,x=-(1+sqrt(15))/2

Find the zeros of 2x^2+4x-9=2(x-1)

The zeros of 2x^2+4x-9=2(x-1) are x=(-1+sqrt(15))/2 ,x=-(1+sqrt(15))/2

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2 x^{2} + 4 x - 9 = 2 (x - 1)

x = \frac{- 1 + 15}{2}, x = - \frac{1 + 15}{2}

Decimal Notation

x = 1.43649 \dots, x = - 2.43649 \dots

Solution steps

2 x^{2} + 4 x - 9 = 2 (x - 1)

Expand

2 (x - 1) : 2 x - 2

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

Move

2

to the left side

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

Move

2 x

to the left side

2 x^{2} + 2 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 7) = 215

x_{1, 2} = \frac{- 2 \pm 2 15}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 15}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 15}{2 \cdot 2}

x = \frac{- 2 + 2 15}{2 \cdot 2} : \frac{- 1 + 15}{2}

x = \frac{- 2 - 2 15}{2 \cdot 2} : - \frac{1 + 15}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 15}{2}, x = - \frac{1 + 15}{2}

\frac{x - 10}{8} = 10

i^{- 2}

- \frac{13}{10} + \frac{1}{5}

2 a - 3 a - 4 - 2 a = 6 a - 2 a - 5 - 4

What are the solutions to the equation 2x^2+4x-9=2(x-1) ?
The solutions to the equation 2x^2+4x-9=2(x-1) are x=(-1+sqrt(15))/2 ,x=-(1+sqrt(15))/2
Find the zeros of 2x^2+4x-9=2(x-1)
The zeros of 2x^2+4x-9=2(x-1) are x=(-1+sqrt(15))/2 ,x=-(1+sqrt(15))/2