Question 195565
{{{15x^2-7x-4=0}}} Start with the given equation.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=15}}}, {{{b=-7}}}, and {{{c=-4}}}



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(15)(-4) ))/(2(15))}}} Plug in  {{{a=15}}}, {{{b=-7}}}, and {{{c=-4}}}



{{{x = (7 +- sqrt( (-7)^2-4(15)(-4) ))/(2(15))}}} Negate {{{-7}}} to get {{{7}}}. 



{{{x = (7 +- sqrt( 49-4(15)(-4) ))/(2(15))}}} Square {{{-7}}} to get {{{49}}}. 



{{{x = (7 +- sqrt( 49--240 ))/(2(15))}}} Multiply {{{4(15)(-4)}}} to get {{{-240}}}



{{{x = (7 +- sqrt( 49+240 ))/(2(15))}}} Rewrite {{{sqrt(49--240)}}} as {{{sqrt(49+240)}}}



{{{x = (7 +- sqrt( 289 ))/(2(15))}}} Add {{{49}}} to {{{240}}} to get {{{289}}}



{{{x = (7 +- sqrt( 289 ))/(30)}}} Multiply {{{2}}} and {{{15}}} to get {{{30}}}. 



{{{x = (7 +- 17)/(30)}}} Take the square root of {{{289}}} to get {{{17}}}. 



{{{x = (7 + 17)/(30)}}} or {{{x = (7 - 17)/(30)}}} Break up the expression. 



{{{x = (24)/(30)}}} or {{{x =  (-10)/(30)}}} Combine like terms. 



{{{x = 4/5}}} or {{{x = -1/3}}} Simplify. 



So the answers are {{{x = 4/5}}} or {{{x = -1/3}}}