Question 183093


{{{2x^2-x=15}}} Start with the given equation.



{{{2x^2-x-15=0}}} Subtract 15 from both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-1}}}, and {{{c=-15}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-1) +- sqrt( (-1)^2-4(2)(-15) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-1}}}, and {{{c=-15}}}



{{{x = (1 +- sqrt( (-1)^2-4(2)(-15) ))/(2(2))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(2)(-15) ))/(2(2))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1--120 ))/(2(2))}}} Multiply {{{4(2)(-15)}}} to get {{{-120}}}



{{{x = (1 +- sqrt( 1+120 ))/(2(2))}}} Rewrite {{{sqrt(1--120)}}} as {{{sqrt(1+120)}}}



{{{x = (1 +- sqrt( 121 ))/(2(2))}}} Add {{{1}}} to {{{120}}} to get {{{121}}}



{{{x = (1 +- sqrt( 121 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (1 +- 11)/(4)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{x = (1 + 11)/(4)}}} or {{{x = (1 - 11)/(4)}}} Break up the expression. 



{{{x = (12)/(4)}}} or {{{x =  (-10)/(4)}}} Combine like terms. 



{{{x = 3}}} or {{{x = -5/2}}} Simplify. 



So the answers are {{{x = 3}}} or {{{x = -5/2}}}