Question 577108


{{{2x^2-16x=-30}}} Start with the given equation.



{{{2x^2-16x+30=0}}} Add 30 to both sides



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=-16}}}, and {{{c=30}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-16) +- sqrt( (-16)^2-4(2)(30) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=-16}}}, and {{{c=30}}}



{{{x = (16 +- sqrt( (-16)^2-4(2)(30) ))/(2(2))}}} Negate {{{-16}}} to get {{{16}}}. 



{{{x = (16 +- sqrt( 256-4(2)(30) ))/(2(2))}}} Square {{{-16}}} to get {{{256}}}. 



{{{x = (16 +- sqrt( 256-240 ))/(2(2))}}} Multiply {{{4(2)(30)}}} to get {{{240}}}



{{{x = (16 +- sqrt( 16 ))/(2(2))}}} Subtract {{{240}}} from {{{256}}} to get {{{16}}}



{{{x = (16 +- sqrt( 16 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (16 +- 4)/(4)}}} Take the square root of {{{16}}} to get {{{4}}}. 



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



{{{x = (20)/(4)}}} or {{{x =  (12)/(4)}}} Combine like terms. 



{{{x = 5}}} or {{{x = 3}}} Simplify. 



So the answers are {{{x = 5}}} or {{{x = 3}}}