How do you solve #x² - 10x = 21#?

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Answer 1 · 2015-07-01T12:27:37

#x^2-10x=21# is a quadratic equation, so start by getting:

#x^2-10x-21=0#

Try to factor, because if it works, it's fast. But in this case, it won't factor nicely. So move on to either completing the square or

For #ax^2+bx+c=0#, the solutions are given by:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

So the solutions for our quadratic equation are:

#x=(-(-10)+-sqrt((-10)^2-4(1)(-21)))/(2(1))#

#=(10+-sqrt(100+84))/2#

#=(10+-sqrt(184))/2 = (10+-sqrt(4*46))/2#

#=(10+-2sqrt46)/2 = (2(5+-sqrt46))/2#

# = 5+-sqrt46#