How do you solve $16 n^{2} - 10 n + 129 = 8 n^{2} - 8$ $16n ^ { 2} - 10n + 129= 8n ^ { 2} - 8$ by completing the square?

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Answer 1 · 2017-03-27T16:22:12

No real solutions

Move the $8 n^{2}$ $8n^2$ to the other side:

$16n^2-10n+129-8n^2=color(red)cancel(color(black)(8n^2))-8color(red)cancel(color(black)(-8n^2))$

$8n^2-10n+129=-8$

Now move 129 to the other side:

$8n^2-10ncolor(red)cancel(color(black)(+129))color(red)cancel(color(black)(-129))=8-129$

$8n^2-10n=-121$

Now add $(b/(2a))^2$ to both sides where $b= -10$ and $a=8$

$8n^2-10n+(-10/16)^2=-121+(-10/16)^2$

$(8n-10/16)^2=-121+(-10/16)^2$

Now if we square root, there will be a negative inside the square root so we will end up getting imaginary solutions. So there are no real solutions

How do you solve 16n ^ { 2} - 10n + 129= 8n ^ { 2} - 816n2−10n+129=8n2−816n ^ { 2} - 10n + 129= 8n ^ { 2} - 8 by completing the square?