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

The solutions to the equation x^2+x=-4 are x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2

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

The zeros of x^2+x=-4 are x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2

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x^{2} + x = - 4

x = - \frac{1}{2} + i \frac{15}{2}, x = - \frac{1}{2} - i \frac{15}{2}

Solution steps

x^{2} + x = - 4

Move

4

to the left side

x^{2} + x + 4 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot 4 : 15 i

x_{1, 2} = \frac{- 1 \pm 15 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 15 i}{2 \cdot 1}, x_{2} = \frac{- 1 - 15 i}{2 \cdot 1}

x = \frac{- 1 + 15 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{15}{2}

x = \frac{- 1 - 15 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{15}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{15}{2}, x = - \frac{1}{2} - i \frac{15}{2}

64 m^{3} + 125

3 (2 x - 14) + x = 15 - (- 9 x - 5)

- \frac{1}{3} y + 10 = - \frac{2}{3} y + 7

- 2 (5 - 4 x) < 6 x - 4

(x^{\frac{1}{2}} \cdot y^{\frac{1}{3}})^{2}

What are the solutions to the equation x^2+x=-4 ?
The solutions to the equation x^2+x=-4 are x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2
Find the zeros of x^2+x=-4
The zeros of x^2+x=-4 are x=-1/2+i(sqrt(15))/2 ,x=-1/2-i(sqrt(15))/2