Question 178400
{{{-x^2+7x-10=0}}} Start with the given equation.



{{{x^2-7x+10=0}}} Multiply EVERY term by -1 to make the leading coefficient positive.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-7}}}, 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 = (-(-7) +- sqrt( (-7)^2-4(1)(10) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-7}}}, and {{{c=10}}}



{{{x = (7 +- sqrt( (-7)^2-4(1)(10) ))/(2(1))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{x = (7 +- sqrt( 49-4(1)(10) ))/(2(1))}}} Square {{{-7}}} to get {{{49}}}. 



{{{x = (7 +- sqrt( 49-40 ))/(2(1))}}} Multiply {{{4(1)(10)}}} to get {{{40}}}



{{{x = (7 +- sqrt( 9 ))/(2(1))}}} Subtract {{{40}}} from {{{49}}} to get {{{9}}}



{{{x = (7 +- sqrt( 9 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (7 +- 3)/(2)}}} Take the square root of {{{9}}} to get {{{3}}}. 



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



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



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



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