Question 115722

{{{2x^2-14x-10=0}}} Start with the given equation



{{{2x^2-14x=10}}} Add 10 to both sides
{{{2(x^2-7x)=10}}} Factor out the leading coefficient 2.  This step is important since we want the {{{x^2}}} coefficient to be equal to 1.




Take half of the x coefficient -7 to {{{-7/2}}}

Now square {{{-7/2}}} to get {{{49/4}}}




{{{2(x^2-7x+49/4)=10+(49/4)(2)}}} Add this result (12.25) to the expression {{{x^2-7x}}}  inside the parenthesis. Now the expression {{{x^2-7x+49/4}}}  is a perfect square trinomial. Now add the result {{{(49/4)(2)}}} (remember we factored out a 2) to the right side.




{{{2(x-7/2)^2=10+(49/4)(2)}}} Factor {{{x^2-7x+49/4}}} into {{{(x-7/2)^2}}} 



{{{2(x-7/2)^2=10+49/2}}} Multiply {{{49/4}}} and 2 to get {{{49/2}}}




{{{2(x-7/2)^2=69/2}}} Combine like terms on the right side


{{{(x-7/2)^2=69/4}}} Divide both sides by 2



{{{x-7/2=0+-sqrt(69/4)}}} Take the square root of both sides



{{{x-7/2=0+-sqrt(69)/2}}} Simplify


{{{x=7/2+-sqrt(69)/2}}} Add {{{7/2}}} to both sides to isolate x.



{{{x=(7+-sqrt(69))/2}}} Combine the fractions