What are the solutions to the equation r^2+4r+196=0 ?

The solutions to the equation r^2+4r+196=0 are r=-2+8sqrt(3)i,r=-2-8sqrt(3)i

Find the zeros of r^2+4r+196=0

The zeros of r^2+4r+196=0 are r=-2+8sqrt(3)i,r=-2-8sqrt(3)i

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r^{2} + 4 r + 196 = 0

r = - 2 + 83 i, r = - 2 - 83 i

Solution steps

r^{2} + 4 r + 196 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 196}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 196 : 163 i

r_{1, 2} = \frac{- 4 \pm 16 3 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 4 + 16 3 i}{2 \cdot 1}, r_{2} = \frac{- 4 - 16 3 i}{2 \cdot 1}

r = \frac{- 4 + 16 3 i}{2 \cdot 1} : - 2 + 83 i

r = \frac{- 4 - 16 3 i}{2 \cdot 1} : - 2 - 83 i

The solutions to the quadratic equation are:

r = - 2 + 83 i, r = - 2 - 83 i

(x + 3) (2 x - 1) = (2 x - 1) (3 x - 5)

1 x^{2} + 6400 x - 115200000 = 0

0 = 150 + 90 t + 4.9 t^{2}

θ^{2} = 4 0^{\circ}

x, y^{2} - x^{2} = k^{2}

What are the solutions to the equation r^2+4r+196=0 ?
The solutions to the equation r^2+4r+196=0 are r=-2+8sqrt(3)i,r=-2-8sqrt(3)i
Find the zeros of r^2+4r+196=0
The zeros of r^2+4r+196=0 are r=-2+8sqrt(3)i,r=-2-8sqrt(3)i