Question 166534
{{{5x^2-3x-3=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=5}}}, {{{b=-3}}}, 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 = (-(-3) +- sqrt( (-3)^2-4(5)(-3) ))/(2(5))}}} Plug in  {{{a=5}}}, {{{b=-3}}}, and {{{c=-3}}}



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



{{{x = (3 +- sqrt( 9-4(5)(-3) ))/(2(5))}}} Square {{{-3}}} to get {{{9}}}. 



{{{x = (3 +- sqrt( 9--60 ))/(2(5))}}} Multiply {{{4(5)(-3)}}} to get {{{-60}}}



{{{x = (3 +- sqrt( 9+60 ))/(2(5))}}} Rewrite {{{sqrt(9--60)}}} as {{{sqrt(9+60)}}}



{{{x = (3 +- sqrt( 69 ))/(2(5))}}} Add {{{9}}} to {{{60}}} to get {{{69}}}



{{{x = (3 +- sqrt( 69 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (3+sqrt(69))/(10)}}} or {{{x = (3-sqrt(69))/(10)}}} Break up the expression.  



So the answers are {{{x = (3+sqrt(69))/(10)}}} or {{{x = (3-sqrt(69))/(10)}}} 



which approximate to {{{x=1.131}}} or {{{x=-0.531}}}