is it possible to solve: 7d^2+12=0??

Yes!

but the solutions will not be real.

\[d^2 = -\frac{12}{7} \]Now take square root on both sides.

If we assume that \(d\) is a real number, then no. If it is allowed to be an imaginary or complex number, then most certainly!

okay. i must've done the previous steps wrong. because the answer is -2 and 2

umm can you solve 10d-2t=-40? then substitute it into 7d^2+5d-t-8=0 pleasse?

d and t?

yeah

WolframAlpha is a good resource to check your answers: http://www.wolframalpha.com/input/?i=10d-2t%3D-40+and+7d%5E2%2B5d-t-8%3D0

Check your work carefully... your troubles were likely caused by an erroneous slip of the pen. (It happens to all of us!)

thanks, but it doesn't show how to solve it algebraically. which I need to do

10d-2t=-40 -2t=-40-10d t = -40/(-2) - 10d/(-2) t = 20+5d <<--- You probably made the error here ------------------- 7d^2+5d-t-8=0 7d^2+5d-(20+5d)-8=0 7d^2+5d-20-5d-8=0 7d^2-28=0 Now solve for d

ohhh ok. yeah i didnt know how to solve for d then I figured it out...thanks!

you're welcome

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