Question 280209
{{{x^2-3x-10=0}}} Start with the given equation.



Notice that the quadratic {{{x^2-3x-10}}} is 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( (-3)^2-4(1)(-10) ))/(2(1))}}} Negate {{{-3}}} to get {{{3}}}. 



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



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



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