Question 118218
{{{x^2-3x-23=5}}} Start with the given equation



{{{x^2-3x-23-5=0}}}  Subtract 5 from both sides. 



{{{x^2-3x-28=0}}} Combine like terms


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 {{{x^2-3*x-28=0}}} ( notice {{{a=1}}}, {{{b=-3}}}, and {{{c=-28}}})





{{{x = (--3 +- sqrt( (-3)^2-4*1*-28 ))/(2*1)}}} Plug in a=1, b=-3, and c=-28




{{{x = (3 +- sqrt( (-3)^2-4*1*-28 ))/(2*1)}}} Negate -3 to get 3




{{{x = (3 +- sqrt( 9-4*1*-28 ))/(2*1)}}} 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+112 ))/(2*1)}}} Multiply {{{-4*-28*1}}} to get {{{112}}}




{{{x = (3 +- sqrt( 121 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (3 +- 11)/(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>)




{{{x = (3 +- 11)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (3 + 11)/2}}} or {{{x = (3 - 11)/2}}}


Lets look at the first part:


{{{x=(3 + 11)/2}}}


{{{x=14/2}}} Add the terms in the numerator

{{{x=7}}} Divide


So one answer is

{{{x=7}}}




Now lets look at the second part:


{{{x=(3 - 11)/2}}}


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

{{{x=-4}}} Divide


So another answer is

{{{x=-4}}}


So our solutions are:

{{{x=7}}} or {{{x=-4}}}


Notice when we graph {{{x^2-3*x-28}}}, we get:


{{{ graph( 500, 500, -14, 17, -14, 17,1*x^2+-3*x+-28) }}}


and we can see that the roots are {{{x=7}}} and {{{x=-4}}}. This verifies our answer