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

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Question 387322: It is possible for a quadratic equation to have no real-number solutions.
Solve
t2 + 10 = 6t
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
t%5E2%2B10=6t Start with the given equation.


t%5E2%2B10-6t=0 Subtract 6t from both sides.


t%5E2-6t%2B10=0 Rearrange the terms.


Notice that the quadratic t%5E2-6t%2B10 is in the form of At%5E2%2BBt%2BC where A=1, B=-6, and C=10


Let's use the quadratic formula to solve for "t":


t+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


t+=+%28-%28-6%29+%2B-+sqrt%28+%28-6%29%5E2-4%281%29%2810%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-6, and C=10


t+=+%286+%2B-+sqrt%28+%28-6%29%5E2-4%281%29%2810%29+%29%29%2F%282%281%29%29 Negate -6 to get 6.


t+=+%286+%2B-+sqrt%28+36-4%281%29%2810%29+%29%29%2F%282%281%29%29 Square -6 to get 36.


t+=+%286+%2B-+sqrt%28+36-40+%29%29%2F%282%281%29%29 Multiply 4%281%29%2810%29 to get 40


t+=+%286+%2B-+sqrt%28+-4+%29%29%2F%282%281%29%29 Subtract 40 from 36 to get -4


t+=+%286+%2B-+sqrt%28+-4+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


t+=+%286+%2B-+2%2Ai%29%2F%282%29 Take the square root of -4 to get 2%2Ai.


t+=+%286+%2B+2%2Ai%29%2F%282%29 or t+=+%286+-+2%2Ai%29%2F%282%29 Break up the expression.


t+=+%286%29%2F%282%29+%2B+%282%2Ai%29%2F%282%29 or t+=++%286%29%2F%282%29+-+%282%2Ai%29%2F%282%29 Break up the fraction for each case.


t+=+3%2Bi or t+=++3-i Reduce.


So the solutions are t+=+3%2Bi or t+=+3-i


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim