Question 449352


{{{5x^2+5=-13x}}} Start with the given equation.



{{{5x^2+13x+5=0}}} Add 13x to both sides.



Notice that the quadratic {{{5x^2+13x+5}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=5}}}, {{{B=13}}}, 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 = (-(13) +- sqrt( (13)^2-4(5)(5) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=13}}}, and {{{C=5}}}



{{{x = (-13 +- sqrt( 169-4(5)(5) ))/(2(5))}}} Square {{{13}}} to get {{{169}}}. 



{{{x = (-13 +- sqrt( 169-100 ))/(2(5))}}} Multiply {{{4(5)(5)}}} to get {{{100}}}



{{{x = (-13 +- sqrt( 69 ))/(2(5))}}} Subtract {{{100}}} from {{{169}}} to get {{{69}}}



{{{x = (-13 +- sqrt( 69 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (-13+sqrt(69))/(10)}}} or {{{x = (-13-sqrt(69))/(10)}}} Break up the expression.  



So the solutions are {{{x = (-13+sqrt(69))/(10)}}} or {{{x = (-13-sqrt(69))/(10)}}}