Question 180712
I'm not sure of the method that you're using, but here's the way that I'd do it:





{{{x^2+3x-10=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=3}}}, 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 = (-(3) +- sqrt( (3)^2-4(1)(-10) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=3}}}, and {{{c=-10}}}



{{{x = (-3 +- sqrt( 9-4(1)(-10) ))/(2(1))}}} Square {{{3}}} to get {{{9}}}. 



{{{x = (-3 +- sqrt( 9--40 ))/(2(1))}}} Multiply {{{4(1)(-10)}}} to get {{{-40}}}



{{{x = (-3 +- sqrt( 9+40 ))/(2(1))}}} Rewrite {{{sqrt(9--40)}}} as {{{sqrt(9+40)}}}



{{{x = (-3 +- sqrt( 49 ))/(2(1))}}} Add {{{9}}} to {{{40}}} to get {{{49}}}



{{{x = (-3 +- sqrt( 49 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-3 +- 7)/(2)}}} Take the square root of {{{49}}} to get {{{7}}}. 



{{{x = (-3 + 7)/(2)}}} or {{{x = (-3 - 7)/(2)}}} Break up the expression. 



{{{x = (4)/(2)}}} or {{{x =  (-10)/(2)}}} Combine like terms. 



{{{x = 2}}} or {{{x = -5}}} Simplify. 



So the answers are {{{x = 2}}} or {{{x = -5}}}