What are the solutions to the equation 5x^2+2x=-4 ?

The solutions to the equation 5x^2+2x=-4 are x=-1/5+i(sqrt(19))/5 ,x=-1/5-i(sqrt(19))/5

Find the zeros of 5x^2+2x=-4

The zeros of 5x^2+2x=-4 are x=-1/5+i(sqrt(19))/5 ,x=-1/5-i(sqrt(19))/5

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

x = - \frac{1}{5} + i \frac{19}{5}, x = - \frac{1}{5} - i \frac{19}{5}

Solution steps

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

Move

4

to the left side

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 \cdot 5 \cdot 4 : 219 i

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

Separate the solutions

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

x = \frac{- 2 + 2 19 i}{2 \cdot 5} : - \frac{1}{5} + i \frac{19}{5}

x = \frac{- 2 - 2 19 i}{2 \cdot 5} : - \frac{1}{5} - i \frac{19}{5}

The solutions to the quadratic equation are:

x = - \frac{1}{5} + i \frac{19}{5}, x = - \frac{1}{5} - i \frac{19}{5}

\frac{2}{2 + k} < 1

6 x^{2} - 2 x = 0

\frac{x}{- 10} \leq - 100

30 + q = 43

- 9 x - 13 = - 103

What are the solutions to the equation 5x^2+2x=-4 ?
The solutions to the equation 5x^2+2x=-4 are x=-1/5+i(sqrt(19))/5 ,x=-1/5-i(sqrt(19))/5
Find the zeros of 5x^2+2x=-4
The zeros of 5x^2+2x=-4 are x=-1/5+i(sqrt(19))/5 ,x=-1/5-i(sqrt(19))/5