How do you use the quadratic formula to solve $tan^2theta+3tantheta+1=0$ for $0<=theta<360$ ?

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Answer 1 · 2016-11-22T19:17:42

$thetaapprox159.095$
$thetaapprox339.095$

$thetaapprox110.905$
$thetaapprox290.905$

Let $u=tantheta$
Rewrite the equation as:
$u^2+3u+1=0$

$u=frac{-(3)+-sqrt((3)^2-4(1)(1))}{2(1)}$
$u=frac{-3+-sqrt5}{2}$

Substitute u back in:
$tantheta=(-3+sqrt5)/2$
$theta=tan^-1((-3+sqrt5)/2)$

$thetaapprox159.095$
$thetaapprox339.095$

$tantheta=(-3-sqrt5)/2$
$theta=tan^-1((-3-sqrt5)/2)$

$thetaapprox110.905$
$thetaapprox290.905$

How do you use the quadratic formula to solve tan^2theta+3tantheta+1=0tan^2theta+3tantheta+1=0 for 0<=theta<3600<=theta<360?