Question 99891
Let's use the quadratic formula to solve for y:



Starting with the general quadratic


{{{ay^2+by+c=0}}}


the general solution using the quadratic equation is:


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




So lets solve {{{3*y^2-7*y-5=0}}} ( notice {{{a=3}}}, {{{b=-7}}}, and {{{c=-5}}})





{{{y = (--7 +- sqrt( (-7)^2-4*3*-5 ))/(2*3)}}} Plug in a=3, b=-7, and c=-5




{{{y = (7 +- sqrt( (-7)^2-4*3*-5 ))/(2*3)}}} Negate -7 to get 7




{{{y = (7 +- sqrt( 49-4*3*-5 ))/(2*3)}}} Square -7 to get 49  (note: remember when you square -7, you must square the negative as well. This is because {{{(-7)^2=-7*-7=49}}}.)




{{{y = (7 +- sqrt( 49+60 ))/(2*3)}}} Multiply {{{-4*-5*3}}} to get {{{60}}}




{{{y = (7 +- sqrt( 109 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{y = (7 +- sqrt(109))/(2*3)}}} 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>)




{{{y = (7 +- sqrt(109))/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{y = (7 + sqrt(109))/6}}} or {{{y = (7 - sqrt(109))/6}}}



Now break up the fraction



{{{y=+7/6+sqrt(109)/6}}} or {{{y=+7/6-sqrt(109)/6}}}



Simplify



{{{y=7 / 6+sqrt(109)/6}}} or {{{y=7 / 6-sqrt(109)/6}}}



So these expressions approximate to


{{{y=2.90671775148509}}} or {{{y=-0.573384418151758}}}



So our solutions are:

{{{y=2.90671775148509}}} or {{{y=-0.573384418151758}}}


Notice when we graph {{{3*x^2-7*x-5}}} (just replace y with x), we get:


{{{ graph( 500, 500, -10.5733844181518, 12.9067177514851, -10.5733844181518, 12.9067177514851,3*x^2+-7*x+-5) }}}


when we use the root finder feature on a calculator, we find that {{{x=2.90671775148509}}} and {{{x=-0.573384418151758}}}.So this verifies our answer