# SOLUTION: It is possible for a quadratic equation to have no real-number solutions. Solve t2 + 10 = 6t

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Question 387441: It is possible for a quadratic equation to have no real-number solutions.
Solve
t2 + 10 = 6t

Found 3 solutions by Alan3354, ptaylor, ewatrrr:
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Yes, it's possible. eg, x^2 + 4 = 0
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t^2 + 10 = 6t
t^2 - 6t + 10 = 0
 Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . The discriminant -4 is less than zero. That means that there are no solutions among real numbers. If you are a student of advanced school algebra and are aware about imaginary numbers, read on. In the field of imaginary numbers, the square root of -4 is + or - . The solution is , or Here's your graph:

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t = 3 ± i
No real number solutions.

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t^2-6t+10=0

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i think so
Hope this helps---ptaylor

You can put this solution on YOUR website!
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Hi,
Yes, when b^2 - 4ac < 0 in the quadratic equaton ax^2 + bx + c = 0
that will result in  being an imaginary number.