Question 242790


{{{w^2-3w+5=0}}} Start with the given equation.



Notice that the quadratic {{{w^2-3w+5}}} is in the form of {{{Aw^2+Bw+C}}} where {{{A=1}}}, {{{B=-3}}}, and {{{C=5}}}



Let's use the quadratic formula to solve for "w":



{{{w = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{w = (-(-3) +- sqrt( (-3)^2-4(1)(5) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-3}}}, and {{{C=5}}}



{{{w = (3 +- sqrt( (-3)^2-4(1)(5) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}. 



{{{w = (3 +- sqrt( 9-4(1)(5) ))/(2(1))}}} Square {{{-3}}} to get {{{9}}}. 



{{{w = (3 +- sqrt( 9-20 ))/(2(1))}}} Multiply {{{4(1)(5)}}} to get {{{20}}}



{{{w = (3 +- sqrt( -11 ))/(2(1))}}} Subtract {{{20}}} from {{{9}}} to get {{{-11}}}



{{{w = (3 +- sqrt( -11 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{w = (3 +- i*sqrt(11))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{w = (3+i*sqrt(11))/(2)}}} or {{{w = (3-i*sqrt(11))/(2)}}} Break up the expression.  



So the solutions are {{{w = (3+i*sqrt(11))/(2)}}} or {{{w = (3-i*sqrt(11))/(2)}}}