Solve 7x^2+8=0?

Do as follows:

7x² + 8 = 0
7x² = -8
x² = -8/7

Now if you take the root of both sides you won't be able to over the real numbers.

But if you know your complex numbers then you will be able to:

x = ± √(-8/7)
x = ± √(-1)(8/7)
x = ± √(-1)√(8/7)
x = ± i√(8/7); because √(-1) = i

Hope this helps!

x² = - 8 / 7
x = ± √( i² (8/7) )
x = ± i √(8/7)

7x^2=-8
x^2=-8/7.
So, that means there is no solution for x in real numbers..

7x^2 + 8 = 0
7x^2 = -8
x^2 = -8/7
x = √(-8/7) (imaginary number)
(no real roots)

In High school, you would just say that this number is invalid.

7x^2 + 8 = 0
7x^2 = -8
x^2 = -8/7
x = √(-8/7) (imaginary number)
(no real roots)

7x^2=-8
x^2=-8/7
Impossible, there is no square root of a negative number. If you want to use imaginary numbers, then the answer is x=i(8/7)^(1/2), if you actually meant 7x^2-8=0, then the answer is x=(8/7)^(1/2)