What are the solutions to the equation r^2+8r+124=0 ?

The solutions to the equation r^2+8r+124=0 are r=-4+6sqrt(3)i,r=-4-6sqrt(3)i

Find the zeros of r^2+8r+124=0

The zeros of r^2+8r+124=0 are r=-4+6sqrt(3)i,r=-4-6sqrt(3)i

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r^{2} + 8 r + 124 = 0

r = - 4 + 63 i, r = - 4 - 63 i

Solution steps

r^{2} + 8 r + 124 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 124}{2 \cdot 1}

Simplify

8^{2} - 4 \cdot 1 \cdot 124 : 123 i

r_{1, 2} = \frac{- 8 \pm 12 3 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 8 + 12 3 i}{2 \cdot 1}, r_{2} = \frac{- 8 - 12 3 i}{2 \cdot 1}

r = \frac{- 8 + 12 3 i}{2 \cdot 1} : - 4 + 63 i

r = \frac{- 8 - 12 3 i}{2 \cdot 1} : - 4 - 63 i

The solutions to the quadratic equation are:

r = - 4 + 63 i, r = - 4 - 63 i

\frac{x ^{2}}{2} = 50.9 - 29, 4 y

- 2 x^{2} = 5

x^{2} + 4 x + \frac{28}{3} = 0

120 z^{2} - 500 z = 0

x (x + 1) = 600

What are the solutions to the equation r^2+8r+124=0 ?
The solutions to the equation r^2+8r+124=0 are r=-4+6sqrt(3)i,r=-4-6sqrt(3)i
Find the zeros of r^2+8r+124=0
The zeros of r^2+8r+124=0 are r=-4+6sqrt(3)i,r=-4-6sqrt(3)i