What are the solutions to the equation r^2+4r+29=0 ?

The solutions to the equation r^2+4r+29=0 are r=-2+5i,r=-2-5i

Find the zeros of r^2+4r+29=0

The zeros of r^2+4r+29=0 are r=-2+5i,r=-2-5i

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r^{2} + 4 r + 29 = 0

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

Solution steps

r^{2} + 4 r + 29 = 0

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 1 \cdot 29 : 10 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

2 (2 x + 3) = x - 13

x^{2} - 17 x + 42

4 r^{2} + r - 3

(\frac{3 \cdot 8.314}{2}) (792 - 298)

\frac{1}{x} \leq 4

What are the solutions to the equation r^2+4r+29=0 ?
The solutions to the equation r^2+4r+29=0 are r=-2+5i,r=-2-5i
Find the zeros of r^2+4r+29=0
The zeros of r^2+4r+29=0 are r=-2+5i,r=-2-5i