What are the solutions to the equation x^2+4x+4=18 ?

The solutions to the equation x^2+4x+4=18 are x=-2+3sqrt(2),x=-2-3sqrt(2)

Find the zeros of x^2+4x+4=18

The zeros of x^2+4x+4=18 are x=-2+3sqrt(2),x=-2-3sqrt(2)

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x^{2} + 4 x + 4 = 18

x = - 2 + 32, x = - 2 - 32

Decimal Notation

x = 2.24264 \dots, x = - 6.24264 \dots

Solution steps

x^{2} + 4 x + 4 = 18

Move

18

to the left side

x^{2} + 4 x - 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 14 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 14) = 62

x_{1, 2} = \frac{- 4 \pm 6 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 6 2}{2 \cdot 1}, x_{2} = \frac{- 4 - 6 2}{2 \cdot 1}

x = \frac{- 4 + 6 2}{2 \cdot 1} : - 2 + 32

x = \frac{- 4 - 6 2}{2 \cdot 1} : - 2 - 32

The solutions to the quadratic equation are:

x = - 2 + 32, x = - 2 - 32

5 d^{2} = 80

3 x - 8 \leq 23 or - 4 x + 26 \geq 6

3 x + 2 (4 + 6 x) = 113

4 w^{2} + 20 w = - 25

49 x^{2} - 1 = 0

What are the solutions to the equation x^2+4x+4=18 ?
The solutions to the equation x^2+4x+4=18 are x=-2+3sqrt(2),x=-2-3sqrt(2)
Find the zeros of x^2+4x+4=18
The zeros of x^2+4x+4=18 are x=-2+3sqrt(2),x=-2-3sqrt(2)