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

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

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

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

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x^{2} + 41 = - 8 x

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

Solution steps

x^{2} + 41 = - 8 x

Move

8 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

s^{2} + 2 s + 1

s^{2} + 4 s + 5

3 x + x^{2} = 10

(y - 15)^{2} = 40 (y + 35)

x^{2} + 5 x + 6

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