Question 215290
{{{x(2x+5)=10}}} Start with the given equation.
 


{{{2x^2+5x=10}}} Distribute



{{{2x^2+5x-10=0}}} Subtract 10 from both sides.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=5}}}, and {{{c=-10}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(5) +- sqrt( (5)^2-4(2)(-10) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=5}}}, and {{{c=-10}}}



{{{x = (-5 +- sqrt( 25-4(2)(-10) ))/(2(2))}}} Square {{{5}}} to get {{{25}}}. 



{{{x = (-5 +- sqrt( 25--80 ))/(2(2))}}} Multiply {{{4(2)(-10)}}} to get {{{-80}}}



{{{x = (-5 +- sqrt( 25+80 ))/(2(2))}}} Rewrite {{{sqrt(25--80)}}} as {{{sqrt(25+80)}}}



{{{x = (-5 +- sqrt( 105 ))/(2(2))}}} Add {{{25}}} to {{{80}}} to get {{{105}}}



{{{x = (-5 +- sqrt( 105 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-5+sqrt(105))/(4)}}} or {{{x = (-5-sqrt(105))/(4)}}} Break up the expression.  



So the answers are {{{x = (-5+sqrt(105))/(4)}}} or {{{x = (-5-sqrt(105))/(4)}}} 



which approximate to {{{x=1.312}}} or {{{x=-3.812}}}