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Mathematics 16 Online
OpenStudy (anonymous):

Solve by completing the square x^2-14x=10

OpenStudy (anonymous):

take half the coefficent of -14, it should be -7. The we have (x-7)^2 which equals=x^2-14x+49. That means you get: (x-7)^2=59

OpenStudy (anonymous):

or (x-7)^2-59=0

jimthompson5910 (jim_thompson5910):

\[\large x^2-14x=10\] \[\large x^2-14x-10=0\] \[\large x^2-14x+49-49-10=0\] \[\large (x-7)^2-49-10=0\] \[\large (x-7)^2-59=0\] \[\large (x-7)^2=59\] \[\large x-7=\pm\sqrt{59}\] \[\large x=7\pm\sqrt{59}\]

OpenStudy (anonymous):

x=7plus or mius sqrt59

OpenStudy (anonymous):

\[x^2-14x=10\] \[(x-7)^2=10+7^2=59\] \[x-7=\pm\sqrt{59}\] \[x=7\pm\sqrt{59}\]

OpenStudy (nilankshi):

good

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