What are the solutions to the equation 2x^2+4x=8 ?

The solutions to the equation 2x^2+4x=8 are x=-1+sqrt(5),x=-1-sqrt(5)

Find the zeros of 2x^2+4x=8

The zeros of 2x^2+4x=8 are x=-1+sqrt(5),x=-1-sqrt(5)

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2 x^{2} + 4 x = 8

x = - 1 + 5, x = - 1 - 5

Decimal Notation

x = 1.23606 \dots, x = - 3.23606 \dots

Solution steps

2 x^{2} + 4 x = 8

Move

8

to the left side

2 x^{2} + 4 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 8 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 8) = 45

x_{1, 2} = \frac{- 4 \pm 4 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 4 5}{2 \cdot 2}, x_{2} = \frac{- 4 - 4 5}{2 \cdot 2}

x = \frac{- 4 + 4 5}{2 \cdot 2} : - 1 + 5

x = \frac{- 4 - 4 5}{2 \cdot 2} : - 1 - 5

The solutions to the quadratic equation are:

x = - 1 + 5, x = - 1 - 5

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

(3 - h)^{2} + (5 - k)^{2} = r^{2}

5 a^{2} + 21 = 49

(2 x - 3)^{2} - 9 = 0

x - 1 = x^{2} - 4 x + 3

What are the solutions to the equation 2x^2+4x=8 ?
The solutions to the equation 2x^2+4x=8 are x=-1+sqrt(5),x=-1-sqrt(5)
Find the zeros of 2x^2+4x=8
The zeros of 2x^2+4x=8 are x=-1+sqrt(5),x=-1-sqrt(5)