What are the solutions to the equation x^2+80=-16x ?

The solutions to the equation x^2+80=-16x are x=-8+4i,x=-8-4i

Find the zeros of x^2+80=-16x

The zeros of x^2+80=-16x are x=-8+4i,x=-8-4i

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x^{2} + 80 = - 16 x

x = - 8 + 4 i, x = - 8 - 4 i

Solution steps

x^{2} + 80 = - 16 x

Move

16 x

to the left side

x^{2} + 80 + 16 x = 0

Write in the standard form

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

x^{2} + 16 x + 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 1 \cdot 80}{2 \cdot 1}

Simplify

1 6^{2} - 4 \cdot 1 \cdot 80 : 8 i

x_{1, 2} = \frac{- 16 \pm 8 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 16 + 8 i}{2 \cdot 1}, x_{2} = \frac{- 16 - 8 i}{2 \cdot 1}

x = \frac{- 16 + 8 i}{2 \cdot 1} : - 8 + 4 i

x = \frac{- 16 - 8 i}{2 \cdot 1} : - 8 - 4 i

The solutions to the quadratic equation are:

x = - 8 + 4 i, x = - 8 - 4 i

\frac{2 3}{12}

m^{8} - n^{8}

(8 \frac{2}{3})^{4}

- 12 \leq 2 x

- 6 y^{3} + 18 y^{4}

What are the solutions to the equation x^2+80=-16x ?
The solutions to the equation x^2+80=-16x are x=-8+4i,x=-8-4i
Find the zeros of x^2+80=-16x
The zeros of x^2+80=-16x are x=-8+4i,x=-8-4i