Question 197253
I'll do the first two to get you started...



# 1


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



Notice we have a quadratic in the form of {{{Aa^2+Ba+C}}} where {{{A=1}}}, {{{B=-4}}}, and {{{C=3}}}



Let's use the quadratic formula to solve for a



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



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



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



{{{a = (4 +- sqrt( 16-4(1)(3) ))/(2(1))}}} Square {{{-4}}} to get {{{16}}}. 



{{{a = (4 +- sqrt( 16-12 ))/(2(1))}}} Multiply {{{4(1)(3)}}} to get {{{12}}}



{{{a = (4 +- sqrt( 4 ))/(2(1))}}} Subtract {{{12}}} from {{{16}}} to get {{{4}}}



{{{a = (4 +- sqrt( 4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{a = (4 +- 2)/(2)}}} Take the square root of {{{4}}} to get {{{2}}}. 



{{{a = (4 + 2)/(2)}}} or {{{a = (4 - 2)/(2)}}} Break up the expression. 



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



{{{a = 3}}} or {{{a = 1}}} Simplify. 



So the solutions are {{{a = 3}}} or {{{a = 1}}} 


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# 2



{{{x^2-4x-10=0}}} Start with the given equation.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-4}}}, and {{{C=-10}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-4) +- sqrt( (-4)^2-4(1)(-10) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-4}}}, and {{{C=-10}}}



{{{x = (4 +- sqrt( (-4)^2-4(1)(-10) ))/(2(1))}}} Negate {{{-4}}} to get {{{4}}}. 



{{{x = (4 +- sqrt( 16-4(1)(-10) ))/(2(1))}}} Square {{{-4}}} to get {{{16}}}. 



{{{x = (4 +- sqrt( 16--40 ))/(2(1))}}} Multiply {{{4(1)(-10)}}} to get {{{-40}}}



{{{x = (4 +- sqrt( 16+40 ))/(2(1))}}} Rewrite {{{sqrt(16--40)}}} as {{{sqrt(16+40)}}}



{{{x = (4 +- sqrt( 56 ))/(2(1))}}} Add {{{16}}} to {{{40}}} to get {{{56}}}



{{{x = (4 +- sqrt( 56 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (4 +- 2*sqrt(14))/(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 = (4)/(2) +- (2*sqrt(14))/(2)}}} Break up the fraction.  



{{{x = 2 +- sqrt(14)}}} Reduce.  



{{{x = 2+sqrt(14)}}} or {{{x = 2-sqrt(14)}}} Break up the expression.  



So the solutions are {{{x = 2+sqrt(14)}}} or {{{x = 2-sqrt(14)}}} 



which approximate to {{{x=5.742}}} or {{{x=-1.742}}}