Question 196505
{{{4x^2-8x+7=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=-8}}}, and {{{c=7}}}



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(4)(7) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=-8}}}, and {{{c=7}}}



{{{x = (8 +- sqrt( (-8)^2-4(4)(7) ))/(2(4))}}} Negate {{{-8}}} to get {{{8}}}. 



{{{x = (8 +- sqrt( 64-4(4)(7) ))/(2(4))}}} Square {{{-8}}} to get {{{64}}}. 



{{{x = (8 +- sqrt( 64-112 ))/(2(4))}}} Multiply {{{4(4)(7)}}} to get {{{112}}}



{{{x = (8 +- sqrt( -48 ))/(2(4))}}} Subtract {{{112}}} from {{{64}}} to get {{{-48}}}



{{{x = (8 +- sqrt( -48 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (8 +- 4i*sqrt(3))/(8)}}} 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 = (8+4i*sqrt(3))/(8)}}} or {{{x = (8-4i*sqrt(3))/(8)}}} Break up the expression.  



{{{x = 8/8+(4i*sqrt(3))/(8)}}} or {{{x = 8/8-(4i*sqrt(3))/(8)}}} Break up the fraction



{{{x = 1+(i*sqrt(3))/(2)}}} or {{{x = 1-(i*sqrt(3))/(2)}}} Reduce




{{{x = 1+(sqrt(3)/2)*i}}} or {{{x = 1-(sqrt(3)/2)*i}}} Rearrange the terms.



So the answers are {{{x = 1+(sqrt(3)/2)*i}}} or {{{x = 1-(sqrt(3)/2)*i}}}



which are in {{{a+bi}}} form where {{{a=1}}} and {{{b=sqrt(3)/2}}}