Question 211101
I'll do the first three to get you going. The rest will be like # 2 and # 3


# 1



{{{(-9+i)/(-9-3i)}}} Start with the given expression.



{{{((-9+i)/(-9-3i))((-9+3i)/(-9+3i))}}} Multiply the fraction by {{{(-9+3i)/(-9+3i)}}}.



Note: {{{-9+3i}}} is the complex conjugate of {{{-9-3i}}}. Multiplying two conjugate pairs will result in a real number.



{{{((-9+i)(-9+3i))/((-9-3i)(-9+3i))}}} Combine the fractions.



{{{((-9)(-9)+(-9)(3i)+(i)(-9)+(i)(3i))/((-9-3i)(-9+3i))}}} FOIL the numerator.



{{{((-9)(-9)+(-9)(3i)+(i)(-9)+(i)(3i))/((-9)(-9)+(-9)(3i)+(-3i)(-9)+(-3i)(3i))}}} FOIL the denominator.



{{{(81-27i-9i+3i^2)/(81-27i+27i-9i^2)}}} Multiply.



{{{(81-27i-9i+3i^2)/(81-27i+27i-9(-1))}}} Replace {{{i^2}}} with -1



{{{(81-27i-9i+3i^2)/(81-27i+27i+9)}}} Multiply



{{{(78-36i)/(90)}}} Combine like terms.



{{{(78)/(90)+((-36)/(90))i}}} Break up the fraction.



{{{13/15-(2/5)i}}} Reduce.



So {{{(-9+i)/(-9-3i)=13/15-(2/5)i}}}.



So the expression is now in standard form {{{a+bi}}} where {{{a=13/15}}} and {{{b=-2/5}}}



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# 2





{{{x^2-6x+17=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-6x+17}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-6}}}, and {{{C=17}}}



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



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



{{{x = (-(-6) +- sqrt( (-6)^2-4(1)(17) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-6}}}, and {{{C=17}}}



{{{x = (6 +- sqrt( (-6)^2-4(1)(17) ))/(2(1))}}} Negate {{{-6}}} to get {{{6}}}. 



{{{x = (6 +- sqrt( 36-4(1)(17) ))/(2(1))}}} Square {{{-6}}} to get {{{36}}}. 



{{{x = (6 +- sqrt( 36-68 ))/(2(1))}}} Multiply {{{4(1)(17)}}} to get {{{68}}}



{{{x = (6 +- sqrt( -32 ))/(2(1))}}} Subtract {{{68}}} from {{{36}}} to get {{{-32}}}



{{{x = (6 +- sqrt( -32 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (6 +- 4i*sqrt(2))/(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 = (6)/(2) +- (4i*sqrt(2))/(2)}}} Break up the fraction.  



{{{x = 3 +- 2*sqrt(2)*i}}} Reduce.  



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



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


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# 3


{{{x^2+3x+5=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+3x+5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=3}}}, and {{{C=5}}}



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)(5) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=3}}}, and {{{C=5}}}



{{{x = (-3 +- sqrt( 9-4(1)(5) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{x = (-3 +- sqrt( 9-20 ))/(2(1))}}} Multiply {{{4(1)(5)}}} to get {{{20}}}



{{{x = (-3 +- sqrt( -11 ))/(2(1))}}} Subtract {{{20}}} from {{{9}}} to get {{{-11}}}



{{{x = (-3 +- sqrt( -11 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-3 +- i*sqrt(11))/(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(11))/(2)}}} or {{{x = (-3-i*sqrt(11))/(2)}}} Break up the expression.  



So the solutions are {{{x = (-3+i*sqrt(11))/(2)}}} or {{{x = (-3-i*sqrt(11))/(2)}}}