Question 317009


{{{x^2-3x=-4}}} Start with the given equation.



{{{x^2-3x+4=0}}} Add 4 to both sides.



Notice that the quadratic {{{x^2-3x+4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-3}}}, and {{{C=4}}}



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



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



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



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



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



{{{x = (3 +- sqrt( 9-16 ))/(2(1))}}} Multiply {{{4(1)(4)}}} to get {{{16}}}



{{{x = (3 +- sqrt( -7 ))/(2(1))}}} Subtract {{{16}}} from {{{9}}} to get {{{-7}}}



{{{x = (3 +- sqrt( -7 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (3 +- i*sqrt(7))/(2)}}} Simplify the square root  



{{{x = (3+i*sqrt(7))/(2)}}} or {{{x = (3-i*sqrt(7))/(2)}}} Break up the expression.  



So the solutions are {{{x = (3+i*sqrt(7))/(2)}}} or {{{x = (3-i*sqrt(7))/(2)}}}