Question

4x^2+25=0

Answer 1

\(4x^2=-25\)

\(x^2=\dfrac{-25}{4}\)

\(x = \pm \sqrt{\dfrac{-25}{4}}\)

Now if you only know about real numbers, you can't have a square root of a negative, so there is so solution for x. If you know about complex nubers, then the solution should be obvious from this, knowing that \(i=\sqrt{-1}\)