Question 120219

{{{3x^2+5x-7=x^2+8x+28}}} Start with the given equation



{{{3x^2+5x-7-x^2-8x-28=0}}}  Subtract x^2 from both sides.  Subtract 8x from both sides.  Subtract 28 from both sides. 



{{{2x^2-3x-35=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 {{{2*x^2-3*x-35=0}}} ( notice {{{a=2}}}, {{{b=-3}}}, and {{{c=-35}}})





{{{x = (--3 +- sqrt( (-3)^2-4*2*-35 ))/(2*2)}}} Plug in a=2, b=-3, and c=-35




{{{x = (3 +- sqrt( (-3)^2-4*2*-35 ))/(2*2)}}} Negate -3 to get 3




{{{x = (3 +- sqrt( 9-4*2*-35 ))/(2*2)}}} 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+280 ))/(2*2)}}} Multiply {{{-4*-35*2}}} to get {{{280}}}




{{{x = (3 +- sqrt( 289 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (3 +- 17)/(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>)




{{{x = (3 +- 17)/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (3 + 17)/4}}} or {{{x = (3 - 17)/4}}}


Lets look at the first part:


{{{x=(3 + 17)/4}}}


{{{x=20/4}}} Add the terms in the numerator

{{{x=5}}} Divide


So one answer is

{{{x=5}}}




Now lets look at the second part:


{{{x=(3 - 17)/4}}}


{{{x=-14/4}}} Subtract the terms in the numerator

{{{x=-7/2}}} Divide


So another answer is

{{{x=-7/2}}}


So our solutions are:

{{{x=5}}} or {{{x=-7/2}}}