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 = 3.23606 \dots, x = - 1.23606 \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

(i - 1)^{2} + \frac{2}{( i - 1 ) ^{2}}

\frac{5}{4} \cdot \frac{1}{2}

4 = x - 10

4 (12 x + 14) \leq 4 - 15 x

49 u^{2} - 28 u + 4

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)