Question 89362
{{{5x^2 + x = 3}}}


{{{5x^2 + x - 3=0}}} Subtract 3 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 solution using the quadratic equation is:


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


So lets solve {{{5*x^2+x-3=0}}} ( notice {{{a=5}}}, {{{b=1}}}, and {{{c=-3}}})


{{{x = (-1 +- sqrt( (1)^2-4*5*-3 ))/(2*5)}}} Plug in a=5, b=1, and c=-3




{{{x = (-1 +- sqrt( 1-4*5*-3 ))/(2*5)}}} Square 1 to get 1  




{{{x = (-1 +- sqrt( 1+60 ))/(2*5)}}} Multiply {{{-4*-3*5}}} to get {{{60}}}




{{{x = (-1 +- sqrt( 61 ))/(2*5)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-1 +- sqrt(61))/(2*5)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{x = (-1 +- sqrt(61))/10}}} Multiply 2 and 5 to get 10


So now the expression breaks down into two parts


{{{x = (-1 + sqrt(61))/10}}} or {{{x = (-1 - sqrt(61))/10}}}



Now break up the fraction



{{{x=-1/10+sqrt(61)/10}}} or {{{x=-1/10-sqrt(61)/10}}}




So these expressions approximate to


{{{x=0.681024967590665}}} or {{{x=-0.881024967590665}}}



So our solutions are:

{{{x=0.681024967590665}}} or {{{x=-0.881024967590665}}}


Notice when we graph {{{5*x^2+x-3}}}, we get:


{{{ graph( 500, 500, -10.8810249675907, 10.6810249675907, -10.8810249675907, 10.6810249675907,5*x^2+1*x+-3) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.681024967590665}}} and {{{x=-0.881024967590665}}}.So this verifies our answer