What are the solutions to the equation x^2-4x+4=12 ?

The solutions to the equation x^2-4x+4=12 are x=2(1+sqrt(3)),x=2(1-sqrt(3))

Find the zeros of x^2-4x+4=12

The zeros of x^2-4x+4=12 are x=2(1+sqrt(3)),x=2(1-sqrt(3))

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x^{2} - 4 x + 4 = 12

x = 2 (1 + 3), x = 2 (1 - 3)

Solution steps

x^{2} - 4 x + 4 = 12

Move

12

to the left side

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 8) = 43

x_{1, 2} = \frac{- ( - 4 ) \pm 4 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 3}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 4 3}{2 \cdot 1}

x = \frac{- ( - 4 ) + 4 3}{2 \cdot 1} : 2 (1 + 3)

x = \frac{- ( - 4 ) - 4 3}{2 \cdot 1} : 2 (1 - 3)

The solutions to the quadratic equation are:

x = 2 (1 + 3), x = 2 (1 - 3)

A = 4 π x^{2}

b^{2} + g = \frac{A}{m}

(8 b - 7)^{2} = 169

x^{2} + 6 x + 10 = 8 x + 25

75 - t^{2} = 0

What are the solutions to the equation x^2-4x+4=12 ?
The solutions to the equation x^2-4x+4=12 are x=2(1+sqrt(3)),x=2(1-sqrt(3))
Find the zeros of x^2-4x+4=12
The zeros of x^2-4x+4=12 are x=2(1+sqrt(3)),x=2(1-sqrt(3))