Question 336631


{{{x^2-3x+4=0}}} Start with the given equation.



Notice we have a quadratic equation 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  (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>)  



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



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



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Jim