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t^2 + 10t + 26 =0
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Use quadratic formula. \[-b+-\sqrt { b^2-4ac \over 2a}\]
a= 1 b=10 c=26
Good job @order
t= \[-10 \pm \sqrt{-2}\] t = \[-10 + i \sqrt{2}\] Or t = \[-10 - i \sqrt{2}\]
What is this formula for @order ?
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It's the same, but I forgot everyything has to go over 2a,
including the -b
no it's not the same b is seperate and has minus sign not over 2a
Ok, but the -b has to go over 2a, and I can't seem to write that in this equation stuffs... :(
No it doesn't have to go over 2a the first formula you wrote before was perfect and right
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I know what you mean
\[ x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}, \]
Yes that's perfect!
yeah :D
Sorry I misunderstood you
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