Question 110771
{{{2x^2=7x+11}}} Start with the given equation


{{{0=-2x^2+7x+11}}} Subtract {{{2x^2}}} from both sides



Let's use the quadratic formula to solve for x:



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{-2*x^2+7*x+11=0}}} ( notice {{{a=-2}}}, {{{b=7}}}, and {{{c=11}}})





{{{x = (-7 +- sqrt( (7)^2-4*-2*11 ))/(2*-2)}}} Plug in a=-2, b=7, and c=11




{{{x = (-7 +- sqrt( 49-4*-2*11 ))/(2*-2)}}} Square 7 to get 49  




{{{x = (-7 +- sqrt( 49+88 ))/(2*-2)}}} Multiply {{{-4*11*-2}}} to get {{{88}}}




{{{x = (-7 +- sqrt( 137 ))/(2*-2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-7 +- sqrt(137))/-4}}} Multiply 2 and -2 to get -4


So now the expression breaks down into two parts


{{{x = (-7 + sqrt(137))/-4}}} or {{{x = (-7 - sqrt(137))/-4}}}



Now break up the fraction



{{{x=-7/-4+sqrt(137)/-4}}} or {{{x=-7/-4-sqrt(137)/-4}}}



Simplify



{{{x=7 / 4-sqrt(137)/4}}} or {{{x=7 / 4+sqrt(137)/4}}}



So these expressions approximate to


{{{x=-1.17617497767991}}} or {{{x=4.67617497767991}}}



So our solutions are:

{{{x=-1.17617497767991}}} or {{{x=4.67617497767991}}}


Notice when we graph {{{-2*x^2+7*x+11}}}, we get:


{{{ graph( 500, 500, -11.1761749776799, 14.6761749776799, -11.1761749776799, 14.6761749776799,-2*x^2+7*x+11) }}}


when we use the root finder feature on a calculator, we find that {{{x=-1.17617497767991}}} and {{{x=4.67617497767991}}}.So this verifies our answer