How do you solve $-2x^2-80=0$ ?

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Answer 1 · 2016-11-08T19:35:16

$x=+-2sqrt(10)i$

$-2x^2-80=0$

$rarr x^2+40=0$

$rarr x^2=-40= i^2 * 2^2 * 10$

$rarr x = +- i2sqrt(10) or +-2sqrt(10)i$

Answer 2 · 2016-11-08T19:40:11

$-2x^2 -80 =0$

Usually you would want a quadratic equation to be equal to 0, but in this case there is no $x$ term, so we will solve it by another method.

Re-arrange to give:

$-2x^2 = 80" "larr div -2$

$x^2 = -40$

$x = +-sqrt(-40)$

There are no real solutions to this equation.

How do you solve -2x^2-80=0-2x^2-80=0?