Question 884370


{{{5x^2-7x-3=0}}} Start with the given equation.



Notice that the quadratic {{{5x^2-7x-3}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=5}}}, {{{B=-7}}}, and {{{C=-3}}}



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(5)(-3) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=-7}}}, and {{{C=-3}}}



{{{x = (7 +- sqrt( (-7)^2-4(5)(-3) ))/(2(5))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{x = (7 +- sqrt( 49-4(5)(-3) ))/(2(5))}}} Square {{{-7}}} to get {{{49}}}. 



{{{x = (7 +- sqrt( 49--60 ))/(2(5))}}} Multiply {{{4(5)(-3)}}} to get {{{-60}}}



{{{x = (7 +- sqrt( 49+60 ))/(2(5))}}} Rewrite {{{sqrt(49--60)}}} as {{{sqrt(49+60)}}}



{{{x = (7 +- sqrt( 109 ))/(2(5))}}} Add {{{49}}} to {{{60}}} to get {{{109}}}



{{{x = (7 +- sqrt( 109 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (7+sqrt(109))/(10)}}} or {{{x = (7-sqrt(109))/(10)}}} Break up the expression.  



So the exact solutions are {{{x = (7+sqrt(109))/(10)}}} or {{{x = (7-sqrt(109))/(10)}}} 



Use a calculator to get these approximate solutions (note: you evaluate the exact solutions to get the approximate solutions)



{{{x = 1.74403065089106}}} or {{{x = -0.34403065089106}}}