What are the solutions to the equation r^2+10r+29=0 ?

The solutions to the equation r^2+10r+29=0 are r=-5+2i,r=-5-2i

Find the zeros of r^2+10r+29=0

The zeros of r^2+10r+29=0 are r=-5+2i,r=-5-2i

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r^{2} + 10 r + 29 = 0

r = - 5 + 2 i, r = - 5 - 2 i

Solution steps

r^{2} + 10 r + 29 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

Simplify

1 0^{2} - 4 \cdot 1 \cdot 29 : 4 i

r_{1, 2} = \frac{- 10 \pm 4 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 10 + 4 i}{2 \cdot 1}, r_{2} = \frac{- 10 - 4 i}{2 \cdot 1}

r = \frac{- 10 + 4 i}{2 \cdot 1} : - 5 + 2 i

r = \frac{- 10 - 4 i}{2 \cdot 1} : - 5 - 2 i

The solutions to the quadratic equation are:

r = - 5 + 2 i, r = - 5 - 2 i

\frac{3 x}{2} - 10 = \frac{1}{2}

3 x^{2} + x - 5 = 0

x^{2} - 8 x - 13 = 0

∣ - 6 - 9 i ∣

- 3 x + 2 > 2 x + 3

What are the solutions to the equation r^2+10r+29=0 ?
The solutions to the equation r^2+10r+29=0 are r=-5+2i,r=-5-2i
Find the zeros of r^2+10r+29=0
The zeros of r^2+10r+29=0 are r=-5+2i,r=-5-2i