Question 263181


{{{x^2-8x+15=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-8x+15}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-8}}}, and {{{C=15}}}



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



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



{{{x = (-(-8) +- sqrt( (-8)^2-4(1)(15) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-8}}}, and {{{C=15}}}



{{{x = (8 +- sqrt( (-8)^2-4(1)(15) ))/(2(1))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{x = (8 +- sqrt( 64-4(1)(15) ))/(2(1))}}} Square {{{-8}}} to get {{{64}}}. 



{{{x = (8 +- sqrt( 64-60 ))/(2(1))}}} Multiply {{{4(1)(15)}}} to get {{{60}}}



{{{x = (8 +- sqrt( 4 ))/(2(1))}}} Subtract {{{60}}} from {{{64}}} to get {{{4}}}



{{{x = (8 +- sqrt( 4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (8 +- 2)/(2)}}} Take the square root of {{{4}}} to get {{{2}}}. 



{{{x = (8 + 2)/(2)}}} or {{{x = (8 - 2)/(2)}}} Break up the expression. 



{{{x = (10)/(2)}}} or {{{x =  (6)/(2)}}} Combine like terms. 



{{{x = 5}}} or {{{x = 3}}} Simplify. 



So the solutions are {{{x = 5}}} or {{{x = 3}}}