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

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

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

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

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

x = - 4 + 5 i, x = - 4 - 5 i

Solution steps

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

Solve with the quadratic formula

x_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 4 + 5 i, x = - 4 - 5 i

5 a + 21 b c + 15 b + 7 a c

\frac{- 9}{- 4}

6 x - 8 = 16

135 (\frac{π}{180})

3 ∣ x + 5 ∣ - 3 < 6

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