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never [62]
1 year ago
3

Can you please completely factor

rmula1" title="11 {x}^{2} + 24x + 1320" alt="11 {x}^{2} + 24x + 1320" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
andriy [413]1 year ago
7 0
\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}

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Answer:

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Step-by-step explanation:

Given equations are:

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Now we have to put the value of y in one of the equation to find the value of x

Putting y = 104 in the first equation

x+y = 152\\x+ 104 = 152\\x = 152-104\\x = 48

Hence,

The solution of the system of equations is x = 48 and y = 104

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