What are the solutions to the equation r^2+8r+41=0 ?

The solutions to the equation r^2+8r+41=0 are r=-4+5i,r=-4-5i

Find the zeros of r^2+8r+41=0

The zeros of r^2+8r+41=0 are r=-4+5i,r=-4-5i

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r^{2} + 8 r + 41 = 0

r = - 4 + 5 i, r = - 4 - 5 i

Solution steps

r^{2} + 8 r + 41 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 41}{2 \cdot 1}

Simplify

8^{2} - 4 \cdot 1 \cdot 41 : 10 i

r_{1, 2} = \frac{- 8 \pm 10 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 8 + 10 i}{2 \cdot 1}, r_{2} = \frac{- 8 - 10 i}{2 \cdot 1}

r = \frac{- 8 + 10 i}{2 \cdot 1} : - 4 + 5 i

r = \frac{- 8 - 10 i}{2 \cdot 1} : - 4 - 5 i

The solutions to the quadratic equation are:

r = - 4 + 5 i, r = - 4 - 5 i

\frac{x}{3} - 8 = - 3

\frac{x}{5} - 11 \geq - 14

6 (2) + 4

2 x^{3} y - x^{2} y + 5 x y^{2} + x y^{3}

\frac{16 !}{( 16 - 6 ) ! 6 !} 0. 5^{6} 0. 5^{16 - 6}

What are the solutions to the equation r^2+8r+41=0 ?
The solutions to the equation r^2+8r+41=0 are r=-4+5i,r=-4-5i
Find the zeros of r^2+8r+41=0
The zeros of r^2+8r+41=0 are r=-4+5i,r=-4-5i