Question 195719
{{{b^2-7b+12=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{Ab^2+Bb+C}}} where {{{A=1}}}, {{{B=-7}}}, and {{{C=12}}}



Let's use the quadratic formula to solve for b



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



{{{b = (-(-7) +- sqrt( (-7)^2-4(1)(12) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-7}}}, and {{{C=12}}}



{{{b = (7 +- sqrt( (-7)^2-4(1)(12) ))/(2(1))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{b = (7 +- sqrt( 49-4(1)(12) ))/(2(1))}}} Square {{{-7}}} to get {{{49}}}. 



{{{b = (7 +- sqrt( 49-48 ))/(2(1))}}} Multiply {{{4(1)(12)}}} to get {{{48}}}



{{{b = (7 +- sqrt( 1 ))/(2(1))}}} Subtract {{{48}}} from {{{49}}} to get {{{1}}}



{{{b = (7 +- sqrt( 1 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{b = (7 +- 1)/(2)}}} Take the square root of {{{1}}} to get {{{1}}}. 



{{{b = (7 + 1)/(2)}}} or {{{b = (7 - 1)/(2)}}} Break up the expression. 



{{{b = (8)/(2)}}} or {{{b =  (6)/(2)}}} Combine like terms. 



{{{b = 4}}} or {{{b = 3}}} Simplify. 



So the answers are {{{b = 4}}} or {{{b = 3}}}