What are the solutions to the equation r^2+9r+65=-r ?

The solutions to the equation r^2+9r+65=-r are r=-5+2sqrt(10)i,r=-5-2sqrt(10)i

Find the zeros of r^2+9r+65=-r

The zeros of r^2+9r+65=-r are r=-5+2sqrt(10)i,r=-5-2sqrt(10)i

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r^{2} + 9 r + 65 = - r

r = - 5 + 210 i, r = - 5 - 210 i

Solution steps

r^{2} + 9 r + 65 = - r

Move

r

to the left side

r^{2} + 10 r + 65 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 65}{2 \cdot 1}

Simplify

1 0^{2} - 4 \cdot 1 \cdot 65 : 410 i

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

Separate the solutions

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

r = \frac{- 10 + 4 10 i}{2 \cdot 1} : - 5 + 210 i

r = \frac{- 10 - 4 10 i}{2 \cdot 1} : - 5 - 210 i

The solutions to the quadratic equation are:

r = - 5 + 210 i, r = - 5 - 210 i

b^{2} + 8 = 33

2 (2 x + 1) = x (x + 5)

x^{2} + (x - 1)^{2} = 5

2 \cdot (2 x + 1) = x \cdot (x + 5)

2 z^{2} - 16 z = - 160

What are the solutions to the equation r^2+9r+65=-r ?
The solutions to the equation r^2+9r+65=-r are r=-5+2sqrt(10)i,r=-5-2sqrt(10)i
Find the zeros of r^2+9r+65=-r
The zeros of r^2+9r+65=-r are r=-5+2sqrt(10)i,r=-5-2sqrt(10)i