Question 152916


{{{x^2-3x=7x-2}}} Start with the given equation.



{{{x^2-3x-7x+2=0}}} Subtract {{{7x}}} from both sides. Add 2 to both sides.



{{{x^2-10x+2=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=1}}}, {{{b=-10}}}, and {{{c=2}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(-10) +- sqrt( (-10)^2-4(1)(2) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=-10}}}, and {{{c=2}}}



{{{x = (10 +- sqrt( (-10)^2-4(1)(2) ))/(2(1))}}} Negate {{{-10}}} to get {{{10}}}. 



{{{x = (10 +- sqrt( 100-4(1)(2) ))/(2(1))}}} Square {{{-10}}} to get {{{100}}}. 



{{{x = (10 +- sqrt( 100-8 ))/(2(1))}}} Multiply {{{4(1)(2)}}} to get {{{8}}}



{{{x = (10 +- sqrt( 92 ))/(2(1))}}} Subtract {{{8}}} from {{{100}}} to get {{{92}}}



{{{x = (10 +- sqrt( 92 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (10 +- 2*sqrt(23))/(2)}}} 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 = (10)/(2) +- (2*sqrt(23))/(2)}}} Break up the fraction.  



{{{x = 5 +- sqrt(23)}}} Reduce.  



{{{x = 5+sqrt(23)}}} or {{{x = 5-sqrt(23)}}} Break up the expression.  



So the solutions are {{{x = 5+sqrt(23)}}} or {{{x = 5-sqrt(23)}}} 



which approximate to {{{x=9.796}}} or {{{x=0.204}}}