Question 96250
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 {{{y^2-2*y-24=0}}} ( notice {{{a=1}}}, {{{b=-2}}}, and {{{c=-24}}})


{{{y = (--2 +- sqrt( (-2)^2-4*1*-24 ))/(2*1)}}} Plug in a=1, b=-2, and c=-24




{{{y = (2 +- sqrt( (-2)^2-4*1*-24 ))/(2*1)}}} Negate -2 to get 2




{{{y = (2 +- sqrt( 4-4*1*-24 ))/(2*1)}}} Square -2 to get 4  (note: remember when you square -2, you must square the negative as well. This is because {{{(-2)^2=-2*-2=4}}}.)




{{{y = (2 +- sqrt( 4+96 ))/(2*1)}}} Multiply {{{-4*-24*1}}} to get {{{96}}}




{{{y = (2 +- sqrt( 100 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{y = (2 +- 10)/(2*1)}}} 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 = (2 +- 10)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{y = (2 + 10)/2}}} or {{{y = (2 - 10)/2}}}


Lets look at the first part:


{{{x=(2 + 10)/2}}}


{{{y=12/2}}} Add the terms in the numerator

{{{y=6}}} Divide


So one answer is

{{{y=6}}}




Now lets look at the second part:


{{{x=(2 - 10)/2}}}


{{{y=-8/2}}} Subtract the terms in the numerator

{{{y=-4}}} Divide


So another answer is

{{{y=-4}}}


So our solutions are:

{{{y=6}}} or {{{y=-4}}}


Notice when we graph {{{x^2-2*x-24}}} (just replace y with x), we get:


{{{ graph( 500, 500, -14, 16, -14, 16,1*x^2+-2*x+-24) }}}


and we can see that the roots are {{{y=6}}} and {{{y=-4}}}. This verifies our answer