What are the solutions to the equation r^2+8r+19=0 ?

The solutions to the equation r^2+8r+19=0 are r=-4+sqrt(3)i,r=-4-sqrt(3)i

Find the zeros of r^2+8r+19=0

The zeros of r^2+8r+19=0 are r=-4+sqrt(3)i,r=-4-sqrt(3)i

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r^{2} + 8 r + 19 = 0

r = - 4 + 3 i, r = - 4 - 3 i

Solution steps

r^{2} + 8 r + 19 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 19}{2 \cdot 1}

Simplify

8^{2} - 4 \cdot 1 \cdot 19 : 23 i

r_{1, 2} = \frac{- 8 \pm 2 3 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 8 + 2 3 i}{2 \cdot 1}, r_{2} = \frac{- 8 - 2 3 i}{2 \cdot 1}

r = \frac{- 8 + 2 3 i}{2 \cdot 1} : - 4 + 3 i

r = \frac{- 8 - 2 3 i}{2 \cdot 1} : - 4 - 3 i

The solutions to the quadratic equation are:

r = - 4 + 3 i, r = - 4 - 3 i

4 (- \frac{5}{4}) + 9 = c (- \frac{5}{4}) + 8

2 (2 x - 3) = 3 (2 x - 4)

9 \cdot \frac{1 - ( 8 - 4 \cdot 3 ) + 9 - 4 \cdot 3}{8 - 3 \cdot 2}

∣ x - 7 ∣ \leq 6

r^{2} - 5 r + 6

What are the solutions to the equation r^2+8r+19=0 ?
The solutions to the equation r^2+8r+19=0 are r=-4+sqrt(3)i,r=-4-sqrt(3)i
Find the zeros of r^2+8r+19=0
The zeros of r^2+8r+19=0 are r=-4+sqrt(3)i,r=-4-sqrt(3)i