How do you solve #5t ^ { 2} + 6= 11t#?

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Answer 1 · 2017-04-22T03:45:38

#t=6/5#
#t=1#

#5t^2+6=11t#

Move all terms to one side, so that the other side is zero:
#5t^2-11t+6=0#

Solve like a quadratic; one way is to solve by factoring:
#5t^2-5t-6t+6=0#

#5t(t-1)-6(t-1)=0#

#color(blue)((5t-6))color(red)((t-1))=0#

Using the zero product rule :
#color(blue)(5t-6)=0#
#t=6/5#

#color(red)(t-1)=0#
#t=1#