What are the solutions to the equation 4x+x^2+16=-4x^2 ?

The solutions to the equation 4x+x^2+16=-4x^2 are x=-2/5+i(2sqrt(19))/5 ,x=-2/5-i(2sqrt(19))/5

Find the zeros of 4x+x^2+16=-4x^2

The zeros of 4x+x^2+16=-4x^2 are x=-2/5+i(2sqrt(19))/5 ,x=-2/5-i(2sqrt(19))/5

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4 x + x^{2} + 16 = - 4 x^{2}

x = - \frac{2}{5} + i \frac{2 19}{5}, x = - \frac{2}{5} - i \frac{2 19}{5}

Solution steps

4 x + x^{2} + 16 = - 4 x^{2}

Move

4 x^{2}

to the left side

5 x^{2} + 4 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 5 \cdot 16}{2 \cdot 5}

Simplify

4^{2} - 4 \cdot 5 \cdot 16 : 419 i

x_{1, 2} = \frac{- 4 \pm 4 19 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 4 + 4 19 i}{2 \cdot 5}, x_{2} = \frac{- 4 - 4 19 i}{2 \cdot 5}

x = \frac{- 4 + 4 19 i}{2 \cdot 5} : - \frac{2}{5} + i \frac{2 19}{5}

x = \frac{- 4 - 4 19 i}{2 \cdot 5} : - \frac{2}{5} - i \frac{2 19}{5}

The solutions to the quadratic equation are:

x = - \frac{2}{5} + i \frac{2 19}{5}, x = - \frac{2}{5} - i \frac{2 19}{5}

x^{2} - 3.2 x + 1.35 = 0

x, 3 x^{2} - 27 = 0

x, f = x^{2} + 2 y

(y - 4)^{2} = 8 (y - 2)

m^{2} - 14 m - 239 = 0

What are the solutions to the equation 4x+x^2+16=-4x^2 ?
The solutions to the equation 4x+x^2+16=-4x^2 are x=-2/5+i(2sqrt(19))/5 ,x=-2/5-i(2sqrt(19))/5
Find the zeros of 4x+x^2+16=-4x^2
The zeros of 4x+x^2+16=-4x^2 are x=-2/5+i(2sqrt(19))/5 ,x=-2/5-i(2sqrt(19))/5