Question 172648
First of all, {{{x^2+2x-3}}} is NOT an equation (as it has no equal sign). Are you trying to solve {{{x^2+2x-3=0}}}





{{{x^2+2x-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=2}}}, 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 = (-(2) +- sqrt( (2)^2-4(1)(-3) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=2}}}, and {{{c=-3}}}



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



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



{{{x = (-2 +- sqrt( 4+12 ))/(2(1))}}} Rewrite {{{sqrt(4--12)}}} as {{{sqrt(4+12)}}}



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



{{{x = (-2 +- sqrt( 16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-2 +- 4)/(2)}}} Take the square root of {{{16}}} to get {{{4}}}. 



{{{x = (-2 + 4)/(2)}}} or {{{x = (-2 - 4)/(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}}}