What are the solutions to the equation r^2+12r+52=0 ?

The solutions to the equation r^2+12r+52=0 are r=-6+4i,r=-6-4i

Find the zeros of r^2+12r+52=0

The zeros of r^2+12r+52=0 are r=-6+4i,r=-6-4i

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r^{2} + 12 r + 52 = 0

r = - 6 + 4 i, r = - 6 - 4 i

Solution steps

r^{2} + 12 r + 52 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 52}{2 \cdot 1}

Simplify

1 2^{2} - 4 \cdot 1 \cdot 52 : 8 i

r_{1, 2} = \frac{- 12 \pm 8 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 12 + 8 i}{2 \cdot 1}, r_{2} = \frac{- 12 - 8 i}{2 \cdot 1}

r = \frac{- 12 + 8 i}{2 \cdot 1} : - 6 + 4 i

r = \frac{- 12 - 8 i}{2 \cdot 1} : - 6 - 4 i

The solutions to the quadratic equation are:

r = - 6 + 4 i, r = - 6 - 4 i

6 \cdot 6 \cdot 6 \cdot 6

- 4 u - 25 < 2 u + 17

16 a^{2} b^{4} + 20 a b^{2} - 6

\frac{a}{b} + \frac{b}{a}

8 y - 3 > 10 or 8 y - 3 < 10

What are the solutions to the equation r^2+12r+52=0 ?
The solutions to the equation r^2+12r+52=0 are r=-6+4i,r=-6-4i
Find the zeros of r^2+12r+52=0
The zeros of r^2+12r+52=0 are r=-6+4i,r=-6-4i