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

The solutions to the equation 0=2x^2-8x+8 are x=2

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

The zeros of 0=2x^2-8x+8 are x=2

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

x = 2

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

(- 8)^{2} - 4 \cdot 2 \cdot 8 = 0

x_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 2}

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

\frac{- ( - 8 )}{2 \cdot 2} = 2

x = 2

The solution to the quadratic equation is:

x = 2

x, 2 x^{2} = 128

3 x^{2} + 14 x + \frac{53}{2} = 0

9 x^{2} + 76 = 54 x

x, 2 x^{2} - 3 x y - 2 y^{2} = 0

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

What are the solutions to the equation 0=2x^2-8x+8 ?
The solutions to the equation 0=2x^2-8x+8 are x=2
Find the zeros of 0=2x^2-8x+8
The zeros of 0=2x^2-8x+8 are x=2