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

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

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

Decimal Notation

x = 1.46410 \dots, x = - 5.46410 \dots

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 (3 - 1)

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

The solutions to the quadratic equation are:

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

x^{2} + 4 x - 21 \geq 0

3 x^{3} + 7 x^{2} + 4 x

(- 6 a^{2} b) \cdot (\frac{1}{2} ab)

\frac{1}{2} \cdot 72 \cdot (0.05 - 0.03)

lo g_{5} (3) = 2 x

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