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

The solutions to the equation (x-2)^2-4x+8=0 are x=6,x=2

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

The zeros of (x-2)^2-4x+8=0 are x=6,x=2

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

x = 6, x = 2

Solution steps

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

Expand

(x - 2)^{2} - 4 x + 8 : x^{2} - 8 x + 12

x^{2} - 8 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 12 = 4

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

Separate the solutions

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

x = \frac{- ( - 8 ) + 4}{2 \cdot 1} : 6

x = \frac{- ( - 8 ) - 4}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 6, x = 2

f^{2} - 4 f = 21

6 x^{2} - 2 = 7 x

2 x^{2} - 16 x + 16 = - 10 + 16

2.5 x^{2} + 25 x - 240 = 0

m^{2} - 12 m = 0

What are the solutions to the equation (x-2)^2-4x+8=0 ?
The solutions to the equation (x-2)^2-4x+8=0 are x=6,x=2
Find the zeros of (x-2)^2-4x+8=0
The zeros of (x-2)^2-4x+8=0 are x=6,x=2