Question 113992
{{{2y^2-y=6}}} Start with the given equation



{{{2y^2-y-6=0}}} Subtract 6 from both sides


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 {{{2*y^2-y-6=0}}} ( notice {{{a=2}}}, {{{b=-1}}}, and {{{c=-6}}})





{{{y = (--1 +- sqrt( (-1)^2-4*2*-6 ))/(2*2)}}} Plug in a=2, b=-1, and c=-6




{{{y = (1 +- sqrt( (-1)^2-4*2*-6 ))/(2*2)}}} Negate -1 to get 1




{{{y = (1 +- sqrt( 1-4*2*-6 ))/(2*2)}}} Square -1 to get 1  (note: remember when you square -1, you must square the negative as well. This is because {{{(-1)^2=-1*-1=1}}}.)




{{{y = (1 +- sqrt( 1+48 ))/(2*2)}}} Multiply {{{-4*-6*2}}} to get {{{48}}}




{{{y = (1 +- sqrt( 49 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{y = (1 +- 7)/(2*2)}}} 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 = (1 +- 7)/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{y = (1 + 7)/4}}} or {{{y = (1 - 7)/4}}}


Lets look at the first part:


{{{x=(1 + 7)/4}}}


{{{y=8/4}}} Add the terms in the numerator

{{{y=2}}} Divide


So one answer is

{{{y=2}}}




Now lets look at the second part:


{{{x=(1 - 7)/4}}}


{{{y=-6/4}}} Subtract the terms in the numerator

{{{y=-3/2}}} Divide


So another answer is

{{{y=-3/2}}}


So our solutions are:

{{{y=2}}} or {{{y=-3/2}}}


Notice when we graph {{{2*x^2-x-6}}} (just replace y with x), we get:


{{{ graph( 500, 500, -13, 12, -13, 12,2*x^2+-1*x+-6) }}}


and we can see that the roots are {{{y=2}}} and {{{y=-3/2}}}. This verifies our answer