What are the solutions to the equation x^2+10x+41=0 ?

The solutions to the equation x^2+10x+41=0 are x=-5+4i,x=-5-4i

Find the zeros of x^2+10x+41=0

The zeros of x^2+10x+41=0 are x=-5+4i,x=-5-4i

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x^{2} + 10 x + 41 = 0

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

Solution steps

x^{2} + 10 x + 41 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot 41}{2 \cdot 1}

Simplify

1 0^{2} - 4 \cdot 1 \cdot 41 : 8 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

(10 + 37) (10 - 37)

5 u^{2} = 20 u - 15

x^{2} + 5 x + 4 \leq 0

85

6 x^{2} + 6 = 12 x + 18

What are the solutions to the equation x^2+10x+41=0 ?
The solutions to the equation x^2+10x+41=0 are x=-5+4i,x=-5-4i
Find the zeros of x^2+10x+41=0
The zeros of x^2+10x+41=0 are x=-5+4i,x=-5-4i