Question 280180


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



Notice that the quadratic {{{x^2+2x-4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=2}}}, and {{{C=-4}}}



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)(-4) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=2}}}, and {{{C=-4}}}



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



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



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



{{{x = (-2 +- sqrt( 20 ))/(2(1))}}} Add {{{4}}} to {{{16}}} to get {{{20}}}



{{{x = (-2 +- sqrt( 20 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-2 +- 2*sqrt(5))/(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 = (-2)/(2) +- (2*sqrt(5))/(2)}}} Break up the fraction.  



{{{x = -1 +- sqrt(5)}}} Reduce.  



{{{x = -1+sqrt(5)}}} or {{{x = -1-sqrt(5)}}} Break up the expression.  



So the solutions are {{{x = -1+sqrt(5)}}} or {{{x = -1-sqrt(5)}}}