Question 92079
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^2-3*x-3=0}}} ( notice {{{a=5}}}, {{{b=-3}}}, and {{{c=-3}}})


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




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




{{{x = (3 +- sqrt( 9-4*5*-3 ))/(2*5)}}} Square -3 to get 9  (note: remember when you square -3, you must square the negative as well. This is because {{{(-3)^2=-3*-3=9}}}.)




{{{x = (3 +- sqrt( 9+60 ))/(2*5)}}} Multiply {{{-4*-3*5}}} to get {{{60}}}




{{{x = (3 +- sqrt( 69 ))/(2*5)}}} Combine like terms in the radicand (everything under the square root)







{{{x = (3 +- sqrt(69))/10}}} Multiply 2 and 5 to get 10


So now the expression breaks down into two parts


{{{x = (3 + sqrt(69))/10}}} or {{{x = (3 - sqrt(69))/10}}}



Now break up the fraction



{{{x=+3/10+sqrt(69)/10}}} or {{{x=+3/10-sqrt(69)/10}}}






So these expressions approximate to


{{{x=1.13066238629181}}} or {{{x=-0.530662386291808}}}



So our solutions are:

{{{x=1.13066238629181}}} or {{{x=-0.530662386291808}}}


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


{{{ graph( 500, 500, -10.5306623862918, 11.1306623862918, -10.5306623862918, 11.1306623862918,5*x^2+-3*x+-3) }}}


when we use the root finder feature on a calculator, we find that {{{x=1.13066238629181}}} and {{{x=-0.530662386291808}}}.So this verifies our answer