Question 169986
{{{w^2-5w=14}}} Start with the given equation.



Take half of the "w" coefficient -5 to get {{{-5/2}}}. Now square {{{-5/2}}} to get {{{(-5/2)^2=25/4}}}. Add this value to both sides



{{{w^2-5w+25/4=14+25/4}}} Add {{{25/4}}} to both sides



{{{w^2-5w+25/4=81/4}}} Combine like terms.



{{{4w^2-4(5w)+cross(4)(25/cross(4))=cross(4)(81/cross(4))}}} Multiply EVERY term by the LCD 4 to clear the fractions



{{{4w^2-20w+25=81}}} Multiply and simplify



{{{(2w-5)^2=81}}} Factor {{{4w^2-20w+25}}} to get {{{(2w-5)^2}}}



{{{2w-5=0+-sqrt(81)}}} Take the square root of both sides



{{{2w-5=9}}} or {{{2w-5=-9}}} Break up the "plus/minus" to form two equations



{{{2w-5=sqrt(81)}}} or {{{2w-5=-sqrt(81)}}} Take the square root of 81 to get 9



{{{2w=5+9}}} or {{{2w=5-9}}} Add 5 to both sides



{{{2w=14}}} or {{{2w=-4}}} Combine like terms.



{{{w=14/2}}} or {{{w=-4/2}}} Divide both sides by 2 to isolate "w"



{{{w=7}}} or {{{w=-2}}} Divide



So the answers are {{{w = 7}}} or {{{w = -2}}}