Question 375827
{{{-x^2-3x+5=0}}} Start with the given equation.



{{{x^2+3x-5=0}}} Multiply EVERY term by -1 to make the leading coefficient positive.



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( 9+20 ))/(2(1))}}} Rewrite {{{sqrt(9--20)}}} as {{{sqrt(9+20)}}}



{{{x = (-3 +- sqrt( 29 ))/(2(1))}}} Add {{{9}}} to {{{20}}} to get {{{29}}}



{{{x = (-3 +- sqrt( 29 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-3+sqrt(29))/(2)}}} or {{{x = (-3-sqrt(29))/(2)}}} Break up the expression.  



So the solutions are {{{x = (-3+sqrt(29))/(2)}}} or {{{x = (-3-sqrt(29))/(2)}}} 



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