What are the solutions to the equation 2x^2-8x+4=0 ?

The solutions to the equation 2x^2-8x+4=0 are x=2+sqrt(2),x=2-sqrt(2)

Find the zeros of 2x^2-8x+4=0

The zeros of 2x^2-8x+4=0 are x=2+sqrt(2),x=2-sqrt(2)

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

x = 2 + 2, x = 2 - 2

Solution steps

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

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 2 \cdot 4 = 42

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

Separate the solutions

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

x = \frac{- ( - 8 ) + 4 2}{2 \cdot 2} : 2 + 2

x = \frac{- ( - 8 ) - 4 2}{2 \cdot 2} : 2 - 2

The solutions to the quadratic equation are:

x = 2 + 2, x = 2 - 2

- 3 ∣ p ∣ = - 12

- 4 x - 14 - 9 x = - 13 x - 7 - 6

60 x^{3} + 50 x^{5}

(27) \frac{4}{3}

2 x^{2} + 5 x - 3 \leq 0

What are the solutions to the equation 2x^2-8x+4=0 ?
The solutions to the equation 2x^2-8x+4=0 are x=2+sqrt(2),x=2-sqrt(2)
Find the zeros of 2x^2-8x+4=0
The zeros of 2x^2-8x+4=0 are x=2+sqrt(2),x=2-sqrt(2)