What are the solutions to the equation r^2+r+4=0 ?

The solutions to the equation r^2+r+4=0 are r=-1/2+i(sqrt(15))/2 ,r=-1/2-i(sqrt(15))/2

Find the zeros of r^2+r+4=0

The zeros of r^2+r+4=0 are r=-1/2+i(sqrt(15))/2 ,r=-1/2-i(sqrt(15))/2

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r^{2} + r + 4 = 0

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

Solution steps

r^{2} + r + 4 = 0

Solve with the quadratic formula

r_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

0 = - x^{2} - 6 x - 7

- 10 + 3 x^{2} = 0

80 + y^{2} = 81

- x^{2} - 3 x - 4 = 0

20 - b^{2} = - 5

What are the solutions to the equation r^2+r+4=0 ?
The solutions to the equation r^2+r+4=0 are r=-1/2+i(sqrt(15))/2 ,r=-1/2-i(sqrt(15))/2
Find the zeros of r^2+r+4=0
The zeros of r^2+r+4=0 are r=-1/2+i(sqrt(15))/2 ,r=-1/2-i(sqrt(15))/2