What are the solutions to the equation r^2+8r+448=0 ?

The solutions to the equation r^2+8r+448=0 are r=-4+12sqrt(3)i,r=-4-12sqrt(3)i

Find the zeros of r^2+8r+448=0

The zeros of r^2+8r+448=0 are r=-4+12sqrt(3)i,r=-4-12sqrt(3)i

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r^{2} + 8 r + 448 = 0

r = - 4 + 123 i, r = - 4 - 123 i

Solution steps

r^{2} + 8 r + 448 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 448}{2 \cdot 1}

Simplify

8^{2} - 4 \cdot 1 \cdot 448 : 243 i

r_{1, 2} = \frac{- 8 \pm 24 3 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 8 + 24 3 i}{2 \cdot 1}, r_{2} = \frac{- 8 - 24 3 i}{2 \cdot 1}

r = \frac{- 8 + 24 3 i}{2 \cdot 1} : - 4 + 123 i

r = \frac{- 8 - 24 3 i}{2 \cdot 1} : - 4 - 123 i

The solutions to the quadratic equation are:

r = - 4 + 123 i, r = - 4 - 123 i

10 - x - 9 x^{2} = 0

x, x^{2} - 63 x y - 5 y^{2} = 8

x^{2} = 3 - 2 x

6 y^{2} - 7 y - 3 = 0

x, 5 x^{2} + 7 x y - y^{2} - x + 3 y + 6 = 0

What are the solutions to the equation r^2+8r+448=0 ?
The solutions to the equation r^2+8r+448=0 are r=-4+12sqrt(3)i,r=-4-12sqrt(3)i
Find the zeros of r^2+8r+448=0
The zeros of r^2+8r+448=0 are r=-4+12sqrt(3)i,r=-4-12sqrt(3)i