What are the solutions to the equation 5x^2+19x=-20 ?

The solutions to the equation 5x^2+19x=-20 are x=-19/10+i(sqrt(39))/(10),x=-19/10-i(sqrt(39))/(10)

Find the zeros of 5x^2+19x=-20

The zeros of 5x^2+19x=-20 are x=-19/10+i(sqrt(39))/(10),x=-19/10-i(sqrt(39))/(10)

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5 x^{2} + 19 x = - 20

x = - \frac{19}{10} + i \frac{39}{10}, x = - \frac{19}{10} - i \frac{39}{10}

Solution steps

5 x^{2} + 19 x = - 20

Move

20

to the left side

5 x^{2} + 19 x + 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 5 \cdot 20}{2 \cdot 5}

Simplify

1 9^{2} - 4 \cdot 5 \cdot 20 : 39 i

x_{1, 2} = \frac{- 19 \pm 39 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 19 + 39 i}{2 \cdot 5}, x_{2} = \frac{- 19 - 39 i}{2 \cdot 5}

x = \frac{- 19 + 39 i}{2 \cdot 5} : - \frac{19}{10} + i \frac{39}{10}

x = \frac{- 19 - 39 i}{2 \cdot 5} : - \frac{19}{10} - i \frac{39}{10}

The solutions to the quadratic equation are:

x = - \frac{19}{10} + i \frac{39}{10}, x = - \frac{19}{10} - i \frac{39}{10}

x + 2 = 10

\frac{2 5 ^{x}}{5 ^{x}}

7 x^{2} - 13 x = 2

- 3 + ∣ n - 2 ∣ > 5

25 x^{2} = 1

What are the solutions to the equation 5x^2+19x=-20 ?
The solutions to the equation 5x^2+19x=-20 are x=-19/10+i(sqrt(39))/(10),x=-19/10-i(sqrt(39))/(10)
Find the zeros of 5x^2+19x=-20
The zeros of 5x^2+19x=-20 are x=-19/10+i(sqrt(39))/(10),x=-19/10-i(sqrt(39))/(10)