Question 87887
{{{x^2-14x=51 }}}


Assuming that the coefficient of {{{x^2}}} is 1, you must take HALF of the x coefficient and SQUARE it, adding this quantity to each side of the equation as follows:

{{{x^2 -14x +_____= 51 + ____}}}  Half of -14 is -7, and (-7)^2= 49!
{{{x^2 -14x+49= 51 + 49}}}
{{{(x-7)^2 = 100}}}


Take the square root of each side:
{{{ (x-7) = 0+- sqrt(100) }}}


Add +7 to each side:
{{{x= 7+- 10 }}}
{{{x= 7+10=17}}}  {{{x=7-10=-3}}}


By the way, since this came out to a square root of a perfect square, like 100, it means that the problem could have been solved by factoring!!


R^2  Retired from SCC