How do you factor $t^2 + t +48$ ?

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Answer 1 · 2018-05-08T10:37:27

$t^2+t+48=(t+1/2+sqrt91/2i)(t+1/2-sqrt91/2i)$

$t^2+t+48$

= $t^2+2xxtxx1/2+(1/2)^2-(1/2)^2+48$

= $(t+1/2)^2+191/4$

= $(t+1/2)^2-(sqrt191/2i)^2$ , where $i^2=-1$

= $(t+1/2+sqrt91/2i)(t+1/2-sqrt91/2i)$

How do you factor t^2 + t +48t^2 + t +48?