Given a triangle with a=11, b=14, and °, what is (are) the possible length(s) of c?

Given a triangle with a=11, b=14, and $\mathbf{\alpha = 15^{\circ}}$, what is (are) the possible length(s) of c?

$\small{ \begin{array}{rcll} a^2 &=& b^2+c^2-2bc\cdot \cos{(\alpha)} \\ c^2-2b\cdot \cos{(\alpha)}\cdot c +b^2-a^2 &=& 0 \\\\ \boxed{~ \begin{array}{rcll} Ac^2+Bc+C = 0 \\ c = {-B \pm \sqrt{B^2-4AC} \over 2A} \end{array} ~}\\\\ c^2-2b\cdot \cos{(\alpha)}\cdot c +b^2-a^2 &=& 0 \qquad A=1 \qquad B = -2b\cdot \cos{(\alpha)} \qquad C = b^2-a^2 \\ c &=& {2b\cdot \cos{(\alpha)} \pm \sqrt{(2b\cdot \cos{(\alpha)})^2-4(b^2-a^2)} \over 2} \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{ [ b\cdot \cos{(\alpha)} ]^2-(b^2-a^2)} \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{b^2\cdot[ \cos^2{(\alpha)} -1] + a^2 } \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{b^2\cdot[ 1-\sin^2{(\alpha)} -1] + a^2 } \\ c &=& b\cdot \cos{(\alpha)} \pm \sqrt{ a^2 - b^2\cdot \sin^2{(\alpha)} } \quad a=11 \quad b=14 \quad \alpha = 15^{\circ}\\ c &=& 14\cdot \cos{(15^{\circ})} \pm \sqrt{ 11^2 - 14^2\cdot \sin^2{(15^{\circ})} } \\ c &=& 14\cdot 0.96592582629 \pm \sqrt{ 107.870489571 } \\ c &=& 13.5229615680 \pm 10.3860719028\\\\ c_1 &=& 13.5229615680 + 10.3860719028\\ c_1 &=& 23.9090334709\\\\ c_2 &=& 13.5229615680 - 10.3860719028\\ c_2 &=& 3.13688966521 \end{array} }$

check:

$\begin{array}{rcll} c_1\cdot c_2 &=& (b-a)(b-a+2a)\\ c_1\cdot c_2 &=& (b-a)(b+a)\\ c_1\cdot c_2 &=& b^2-a^2 \quad a=11 \quad b=14 \\ c_1\cdot c_2 &=& 14^2-11^2 \\ c_1\cdot c_2 &=& 75 \\\\ c_1\cdot c_2 &=& 23.9090334709\cdot 3.13688966521 \\ c_1\cdot c_2 &=& 75\quad \text{ Okay}\\ \end{array}$