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16x2+8x=0
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32+8x=0 8x=32 x=-4
\[16x^2\] or 16x2 ?
good question
yes cuz the answer is defferent
that is very true
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if that was \[16^2 \]then the solve is \[16x^2+8x=0\]=\[16x^2=-8x\] divide it x so we have \[16x=-8\]\[x=-0.5\]
ohhh. simple :) first of all, it is the same as: \[2x^2 + x = 0\] then follow whatever Abdullah Al-Hagbani said. :D
If its 16x^2 + 8x = 0 8x(2x+1) = 0 x = 0 or x -1/2
x = -1/2
in finally x = -1/2 .. cuz 16x = -8 then x = -1/2
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