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

The solutions to the equation x^2+4x+4=x-8 are x=-3/2+i(sqrt(39))/2 ,x=-3/2-i(sqrt(39))/2

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

The zeros of x^2+4x+4=x-8 are x=-3/2+i(sqrt(39))/2 ,x=-3/2-i(sqrt(39))/2

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

x = - \frac{3}{2} + i \frac{39}{2}, x = - \frac{3}{2} - i \frac{39}{2}

Solution steps

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

Move

8

to the left side

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

Move

x

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

Simplify

3^{2} - 4 \cdot 1 \cdot 12 : 39 i

x_{1, 2} = \frac{- 3 \pm 39 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 39 i}{2 \cdot 1}, x_{2} = \frac{- 3 - 39 i}{2 \cdot 1}

x = \frac{- 3 + 39 i}{2 \cdot 1} : - \frac{3}{2} + i \frac{39}{2}

x = \frac{- 3 - 39 i}{2 \cdot 1} : - \frac{3}{2} - i \frac{39}{2}

The solutions to the quadratic equation are:

x = - \frac{3}{2} + i \frac{39}{2}, x = - \frac{3}{2} - i \frac{39}{2}

y, s (y) = 7 yy + 9

t (t - 4) = 0

x, 2 x^{2} y^{2} = c_{1}

3 x^{2} - 4 x + 7 x - 18 = 2 x^{2}

x^{2} + 4 x - 12 = 0

What are the solutions to the equation x^2+4x+4=x-8 ?
The solutions to the equation x^2+4x+4=x-8 are x=-3/2+i(sqrt(39))/2 ,x=-3/2-i(sqrt(39))/2
Find the zeros of x^2+4x+4=x-8
The zeros of x^2+4x+4=x-8 are x=-3/2+i(sqrt(39))/2 ,x=-3/2-i(sqrt(39))/2