Question 88257
{{{2x^2 + 5x = 4}}}


{{{2x^2 + 5x - 4=0 }}}Subtract 4 from both sides


Now let's use the quadratic formula to solve for x:



Starting with the general quadratic


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


the general form of the quadratic equation is:


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


So lets solve {{{2*x^2+5*x-4=0}}} (notice {{{a=2}}}, {{{b=5}}}, and {{{c=-4}}})


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




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




{{{x = (-5 +- sqrt( 25+32 ))/(2*2)}}} Multiply {{{-4*-4*2}}} to get {{{32}}}




{{{x = (-5 +- sqrt( 57 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-5 +- sqrt(57))/(2*2)}}} Simplify the square root




{{{x = (-5 +- sqrt(57))/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (-5 + sqrt(57))/4}}} or {{{x = (-5 - sqrt(57))/4}}}



Which approximate to


{{{x=0.637458608817687}}} or {{{x=-3.13745860881769}}}



So our solutions are:

{{{x=0.637458608817687}}} or {{{x=-3.13745860881769}}}


Notice when we graph {{{2*x^2+5*x-4}}} we get:


{{{ graph( 500, 500, -13.1374586088177, 10.6374586088177, -13.1374586088177, 10.6374586088177,2*x^2+5*x+-4) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.637458608817687}}} and {{{x=-3.13745860881769}}}.So this verifies our answer