Question 171774


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



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=4}}}, and {{{c=3}}}



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)(3) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=4}}}, and {{{c=3}}}



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



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



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



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



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



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



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



{{{x = -1}}} or {{{x = -3}}} Simplify. 



So the answers are {{{x = -1}}} or {{{x = -3}}}