What are the solutions to the equation x^2+4x+4=128 ?

The solutions to the equation x^2+4x+4=128 are x=2(4sqrt(2)-1),x=-2(1+4sqrt(2))

Find the zeros of x^2+4x+4=128

The zeros of x^2+4x+4=128 are x=2(4sqrt(2)-1),x=-2(1+4sqrt(2))

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x^{2} + 4 x + 4 = 128

x = 2 (42 - 1), x = - 2 (1 + 42)

Decimal Notation

x = 9.31370 \dots, x = - 13.31370 \dots

Solution steps

x^{2} + 4 x + 4 = 128

Move

128

to the left side

x^{2} + 4 x - 124 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 124 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 124) = 162

x_{1, 2} = \frac{- 4 \pm 16 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 16 2}{2 \cdot 1}, x_{2} = \frac{- 4 - 16 2}{2 \cdot 1}

x = \frac{- 4 + 16 2}{2 \cdot 1} : 2 (42 - 1)

x = \frac{- 4 - 16 2}{2 \cdot 1} : - 2 (1 + 42)

The solutions to the quadratic equation are:

x = 2 (42 - 1), x = - 2 (1 + 42)

0 = x^{2} - 12 x + 24

9 = (x + 5)^{2}

0 = x^{2} - 12 x + 32

0 = 2 x^{2} - x - 10

x^{2} + 2.2 x - 10.35 = 0

What are the solutions to the equation x^2+4x+4=128 ?
The solutions to the equation x^2+4x+4=128 are x=2(4sqrt(2)-1),x=-2(1+4sqrt(2))
Find the zeros of x^2+4x+4=128
The zeros of x^2+4x+4=128 are x=2(4sqrt(2)-1),x=-2(1+4sqrt(2))