What are the solutions to the equation r^2+12r+27=0 ?

The solutions to the equation r^2+12r+27=0 are r=-3,r=-9

Find the zeros of r^2+12r+27=0

The zeros of r^2+12r+27=0 are r=-3,r=-9

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r^{2} + 12 r + 27 = 0

r = - 3, r = - 9

Solution steps

r^{2} + 12 r + 27 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 27}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 27 = 6

r_{1, 2} = \frac{- 12 \pm 6}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 12 + 6}{2 \cdot 1}, r_{2} = \frac{- 12 - 6}{2 \cdot 1}

r = \frac{- 12 + 6}{2 \cdot 1} : - 3

r = \frac{- 12 - 6}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

r = - 3, r = - 9

2 (x + 4) = 12

2 lo g_{5} (x) + lo g_{5} (3)

698

(2 + i) (2 - i)

- 8 \leq 6 x - 2 \leq 4

What are the solutions to the equation r^2+12r+27=0 ?
The solutions to the equation r^2+12r+27=0 are r=-3,r=-9
Find the zeros of r^2+12r+27=0
The zeros of r^2+12r+27=0 are r=-3,r=-9