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

The solutions to the equation x^2-4x+8=4 are x=2

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

The zeros of x^2-4x+8=4 are x=2

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

x = 2

Solution steps

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

Move

4

to the left side

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 4 = 0

x_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 1}

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

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

x = 2

The solution to the quadratic equation is:

x = 2

\frac{16 - y ^{2}}{8} = \frac{y ^{2} - 48}{24}

x, 2 (x^{2} + y^{2}) 2 = 25 (x^{2} - y^{2})

\frac{1}{2} k^{2} + 53 = 5

k^{2} - 4 k = 0

n^{2} - 4 n - 45 = 0

What are the solutions to the equation x^2-4x+8=4 ?
The solutions to the equation x^2-4x+8=4 are x=2
Find the zeros of x^2-4x+8=4
The zeros of x^2-4x+8=4 are x=2