Question

What are the possible values of x in 8x2 + 4x = -1?

Answer 1

What part are you having troubles with, human?

Answer 2

*calculates*

You will get complex solutions

Answer 3

\[4x(2x+1)=-1\]

\[x(2x+1) = -\frac{ 1 }{ 4 }\]

\[2x^2+x = -\frac{ 1 }{ 4 }\]

\[x^2+\frac{ x }{ 2 }+\frac{ 1 }{ 16 }=-\frac{ 1 }{ 16 }\]

\[\left( x+\frac{ 1 }{ 4 } \right)^2=-\frac{ 1 }{ 16 }\]

Answer 4

\[x= -\frac{ 1 }{ 4 }+\frac{ i }{ 4 }\]