Question 189008
{{{8x^2-12x=11}}} Start with the given equation.



{{{8x^2-12x-11=0}}} Subtract 11 from both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=8}}}, {{{b=-12}}}, and {{{c=-11}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-12) +- sqrt( (-12)^2-4(8)(-11) ))/(2(8))}}} Plug in  {{{a=8}}}, {{{b=-12}}}, and {{{c=-11}}}



{{{x = (12 +- sqrt( (-12)^2-4(8)(-11) ))/(2(8))}}} Negate {{{-12}}} to get {{{12}}}. 



{{{x = (12 +- sqrt( 144-4(8)(-11) ))/(2(8))}}} Square {{{-12}}} to get {{{144}}}. 



{{{x = (12 +- sqrt( 144--352 ))/(2(8))}}} Multiply {{{4(8)(-11)}}} to get {{{-352}}}



{{{x = (12 +- sqrt( 144+352 ))/(2(8))}}} Rewrite {{{sqrt(144--352)}}} as {{{sqrt(144+352)}}}



{{{x = (12 +- sqrt( 496 ))/(2(8))}}} Add {{{144}}} to {{{352}}} to get {{{496}}}



{{{x = (12 +- sqrt( 496 ))/(16)}}} Multiply {{{2}}} and {{{8}}} to get {{{16}}}. 



{{{x = (12 +- 4*sqrt(31))/(16)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (12+4*sqrt(31))/(16)}}} or {{{x = (12-4*sqrt(31))/(16)}}} Break up the expression.  



{{{x = (3+sqrt(31))/(4)}}} or {{{x = (3-sqrt(31))/(4)}}} Reduce



So the answers are {{{x = (3+sqrt(31))/(4)}}} or {{{x = (3-sqrt(31))/(4)}}}