Question

Help please?? Fan and medal ~
Solve 2x^2 + 16x + 27 = 0 by completing the square.

Answer 1

\[x^2+8x=-\frac{ 27 }{ 2 }\]
adding both sides \[\left( \frac{ 8 }{ 2 } \right)^2or~16\]
\[x^2+8x+16=16-\frac{ 27 }{ 2 }=\frac{ 5 }{ 2 }\]
\[\left( x+4 \right)^2=\frac{ 5 }{ 2 }\]
\[x+4=\pm \frac{ 5 }{ 2 }\]
\[x=-4 \pm \frac{ 5 }{ 2 }\]
find both values of x

Answer 2

Thank you so much!! I have one more question, mind if I open another on and ask you??

Answer 3

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