can you please completely factor [tex]11 {x}^{2} + 24x + 1320[/tex]

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03-11-2022
Mathematics

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can you please completely factor [tex]11 {x}^{2} + 24x + 1320[/tex]

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TamemC101323 TamemC101323 · Answer 1 · 2022-11-03T13:24:08+01:00

[tex]\begin{gathered} 11x^2+24+1320 \\ ax^2+bx+c=0 \\ \\ a=11\text{ b=24 and c=1320} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-24\pm\sqrt[]{24^2^{}-4\times11\times1320}}{2\times11} \\ \\ x=\frac{-24\pm\sqrt[]{576^{}-58080}}{22} \\ \\ x=\frac{-24\pm\sqrt[]{-57504}}{22} \\ x=\frac{-24\pm\sqrt[]{-57504}}{22} \\ x=\frac{-24\pm i239.8}{22} \\ \\ x=\frac{-24+i239.8}{22}\text{ or }x=\frac{-24-i239.8}{22} \\ (x+\frac{24-i239.8}{22})(x+\frac{24+i239.8}{22}) \end{gathered}[/tex]