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

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

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

The zeros of x^2-4x-4+8=0 are x=2

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

x = 2

Solution steps

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

Refine

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

12 x^{2} + 3 - 12 x = 0

3 (t - 2)^{2} = t (t + 5) + 4

3 x^{2} - 6 x + c = 0

x, f (x + 3) = g (x^{2})

x^{2} + 86 x + 168 = 0

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