What are the solutions to the equation 2x^2+4x=2 ?

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

Find the zeros of 2x^2+4x=2

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

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2 x^{2} + 4 x = 2

x = - 1 + 2, x = - 1 - 2

Decimal Notation

x = 0.41421 \dots, x = - 2.41421 \dots

Solution steps

2 x^{2} + 4 x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

4^{2} - 4 \cdot 2 (- 2) = 42

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

Separate the solutions

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

x = \frac{- 4 + 4 2}{2 \cdot 2} : - 1 + 2

x = \frac{- 4 - 4 2}{2 \cdot 2} : - 1 - 2

The solutions to the quadratic equation are:

x = - 1 + 2, x = - 1 - 2

x^{2} + x = 0

(x^{2} - 10 x + 25) + 6 = 42

\frac{1}{2} x^{2} = - 32

3 = 11 m^{2}

2 x^{2} - 5 x + 25 = 0

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