What are the solutions to the equation r^2+4r+13=0 ?

The solutions to the equation r^2+4r+13=0 are r=-2+3i,r=-2-3i

Find the zeros of r^2+4r+13=0

The zeros of r^2+4r+13=0 are r=-2+3i,r=-2-3i

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r^{2} + 4 r + 13 = 0

r = - 2 + 3 i, r = - 2 - 3 i

Solution steps

r^{2} + 4 r + 13 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 13 : 6 i

r_{1, 2} = \frac{- 4 \pm 6 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 4 + 6 i}{2 \cdot 1}, r_{2} = \frac{- 4 - 6 i}{2 \cdot 1}

r = \frac{- 4 + 6 i}{2 \cdot 1} : - 2 + 3 i

r = \frac{- 4 - 6 i}{2 \cdot 1} : - 2 - 3 i

The solutions to the quadratic equation are:

r = - 2 + 3 i, r = - 2 - 3 i

0.5 x + 78.2 = 287

w^{2} + 5 w + 6

8 x = 0

- 2 x^{- 3}

\frac{9 !}{5 ! ( 9 - 5 ) !}

What are the solutions to the equation r^2+4r+13=0 ?
The solutions to the equation r^2+4r+13=0 are r=-2+3i,r=-2-3i
Find the zeros of r^2+4r+13=0
The zeros of r^2+4r+13=0 are r=-2+3i,r=-2-3i