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:

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

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

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

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

$= 5+-sqrt46$

How do you solve x² - 10x = 21x² - 10x = 21?