Question 158210
I'll start where you left off



{{{2x^2-50x+312=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-50}}}, and {{{c=312}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(-50) +- sqrt( (-50)^2-4(2)(312) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-50}}}, and {{{c=312}}}



{{{x = (50 +- sqrt( (-50)^2-4(2)(312) ))/(2(2))}}} Negate {{{-50}}} to get {{{50}}}. 



{{{x = (50 +- sqrt( 2500-4(2)(312) ))/(2(2))}}} Square {{{-50}}} to get {{{2500}}}. 



{{{x = (50 +- sqrt( 2500-2496 ))/(2(2))}}} Multiply {{{4(2)(312)}}} to get {{{2496}}}



{{{x = (50 +- sqrt( 4 ))/(2(2))}}} Subtract {{{2496}}} from {{{2500}}} to get {{{4}}}



{{{x = (50 +- sqrt( 4 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (50 +- 2)/(4)}}} Take the square root of {{{4}}} to get {{{2}}}. 



{{{x = (50 + 2)/(4)}}} or {{{x = (50 - 2)/(4)}}} Break up the expression. 



{{{x = (52)/(4)}}} or {{{x =  (48)/(4)}}} Combine like terms. 



{{{x = 13}}} or {{{x = 12}}} Simplify. 



So the answers are {{{x = 13}}} or {{{x = 12}}} 

  

Now use these values of x to find the values of y. Let me know if you have any questions.




Note: your answers should be (12,13) or (13,12) which means that the two numbers are 12 and 13