What are the solutions to the equation r^2+2r+32=0 ?

The solutions to the equation r^2+2r+32=0 are r=-1+sqrt(31)i,r=-1-sqrt(31)i

Find the zeros of r^2+2r+32=0

The zeros of r^2+2r+32=0 are r=-1+sqrt(31)i,r=-1-sqrt(31)i

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r^{2} + 2 r + 32 = 0

r = - 1 + 31 i, r = - 1 - 31 i

Solution steps

r^{2} + 2 r + 32 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

Simplify

2^{2} - 4 \cdot 1 \cdot 32 : 231 i

r_{1, 2} = \frac{- 2 \pm 2 31 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 2 + 2 31 i}{2 \cdot 1}, r_{2} = \frac{- 2 - 2 31 i}{2 \cdot 1}

r = \frac{- 2 + 2 31 i}{2 \cdot 1} : - 1 + 31 i

r = \frac{- 2 - 2 31 i}{2 \cdot 1} : - 1 - 31 i

The solutions to the quadratic equation are:

r = - 1 + 31 i, r = - 1 - 31 i

x^{2} + 2 x - 8 = 0

16 = 38 r - 12 r^{2}

5 x^{2} = 36

9 x^{2} - 18 = 0

- 0.0002037 T^{2} + 0.0170086 T - 0.06426 = 0

What are the solutions to the equation r^2+2r+32=0 ?
The solutions to the equation r^2+2r+32=0 are r=-1+sqrt(31)i,r=-1-sqrt(31)i
Find the zeros of r^2+2r+32=0
The zeros of r^2+2r+32=0 are r=-1+sqrt(31)i,r=-1-sqrt(31)i