Question

find zero of the polynomial 4u^2 +8

Answer 1

4u^2+8u=0
4u(u+2)=0
u=0
u=-2

Answer 2

$to \: find \: zeroes \: \\ we \: have \: to \: put \: function \: equals to \: 0 \\ \: 4 {u}^{2} + 8 = 0 \\4 {u}^{2} = - 8 \\ {u}^{2} = - 2 \\ u = + \sqrt{ - 2} and \: - \sqrt{ - 2} \\ \\ + \sqrt{2} i \\ and \: - \sqrt{2} i \\ i = \sqrt{ - 1} \: called \: iota$