What are the solutions to the equation 2x^2+8x-5=2(x-1) ?

The solutions to the equation 2x^2+8x-5=2(x-1) are x=(-3+sqrt(15))/2 ,x=-(3+sqrt(15))/2

Find the zeros of 2x^2+8x-5=2(x-1)

The zeros of 2x^2+8x-5=2(x-1) are x=(-3+sqrt(15))/2 ,x=-(3+sqrt(15))/2

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2 x^{2} + 8 x - 5 = 2 (x - 1)

x = \frac{- 3 + 15}{2}, x = - \frac{3 + 15}{2}

Solution steps

2 x^{2} + 8 x - 5 = 2 (x - 1)

Expand

2 (x - 1) : 2 x - 2

2 x^{2} + 8 x - 5 = 2 x - 2

Move

2

to the left side

2 x^{2} + 8 x - 3 = 2 x

Move

2 x

to the left side

2 x^{2} + 6 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 3) = 215

x_{1, 2} = \frac{- 6 \pm 2 15}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 6 + 2 15}{2 \cdot 2}, x_{2} = \frac{- 6 - 2 15}{2 \cdot 2}

x = \frac{- 6 + 2 15}{2 \cdot 2} : \frac{- 3 + 15}{2}

x = \frac{- 6 - 2 15}{2 \cdot 2} : - \frac{3 + 15}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 15}{2}, x = - \frac{3 + 15}{2}

\frac{m}{15} = 3

\frac{x}{3} - 1 \geq 8

(5 x^{2} - 6) + (- 6 x^{2} - 5 x)

\frac{x}{7} = 3

- 3 \leq x < 1

What are the solutions to the equation 2x^2+8x-5=2(x-1) ?
The solutions to the equation 2x^2+8x-5=2(x-1) are x=(-3+sqrt(15))/2 ,x=-(3+sqrt(15))/2
Find the zeros of 2x^2+8x-5=2(x-1)
The zeros of 2x^2+8x-5=2(x-1) are x=(-3+sqrt(15))/2 ,x=-(3+sqrt(15))/2