What are the solutions to the equation 5x^2+4=-8x ?

The solutions to the equation 5x^2+4=-8x are x=-4/5+i 2/5 ,x=-4/5-i 2/5

Find the zeros of 5x^2+4=-8x

The zeros of 5x^2+4=-8x are x=-4/5+i 2/5 ,x=-4/5-i 2/5

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

x = - \frac{4}{5} + i \frac{2}{5}, x = - \frac{4}{5} - i \frac{2}{5}

Solution steps

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

Move

8 x

to the left side

5 x^{2} + 4 + 8 x = 0

Write in the standard form

a x^{2} + b x + c = 0

5 x^{2} + 8 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 5 \cdot 4}{2 \cdot 5}

Simplify

8^{2} - 4 \cdot 5 \cdot 4 : 4 i

x_{1, 2} = \frac{- 8 \pm 4 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 8 + 4 i}{2 \cdot 5}, x_{2} = \frac{- 8 - 4 i}{2 \cdot 5}

x = \frac{- 8 + 4 i}{2 \cdot 5} : - \frac{4}{5} + i \frac{2}{5}

x = \frac{- 8 - 4 i}{2 \cdot 5} : - \frac{4}{5} - i \frac{2}{5}

The solutions to the quadratic equation are:

x = - \frac{4}{5} + i \frac{2}{5}, x = - \frac{4}{5} - i \frac{2}{5}

5 x + 2 < 5 (x + 1)

x^{2} - 10 x + 27 \leq 0

\frac{10}{6}

- 4 < 2 (x + 1) \leq 12

What are the solutions to the equation 5x^2+4=-8x ?
The solutions to the equation 5x^2+4=-8x are x=-4/5+i 2/5 ,x=-4/5-i 2/5
Find the zeros of 5x^2+4=-8x
The zeros of 5x^2+4=-8x are x=-4/5+i 2/5 ,x=-4/5-i 2/5