Question 124298


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



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



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^2-5*x-3=0}}} ( notice {{{a=2}}}, {{{b=-5}}}, and {{{c=-3}}})





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




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




{{{x = (5 +- sqrt( 25-4*2*-3 ))/(2*2)}}} Square -5 to get 25  (note: remember when you square -5, you must square the negative as well. This is because {{{(-5)^2=-5*-5=25}}}.)




{{{x = (5 +- sqrt( 25+24 ))/(2*2)}}} Multiply {{{-4*-3*2}}} to get {{{24}}}




{{{x = (5 +- sqrt( 49 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (5 +- 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>)




{{{x = (5 +- 7)/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (5 + 7)/4}}} or {{{x = (5 - 7)/4}}}


Lets look at the first part:


{{{x=(5 + 7)/4}}}


{{{x=12/4}}} Add the terms in the numerator

{{{x=3}}} Divide


So one answer is

{{{x=3}}}




Now lets look at the second part:


{{{x=(5 - 7)/4}}}


{{{x=-2/4}}} Subtract the terms in the numerator

{{{x=-1/2}}} Divide


So another answer is

{{{x=-1/2}}}


So our solutions are:

{{{x=3}}} or {{{x=-1/2}}}


Notice when we graph {{{2*x^2-5*x-3}}}, we get:


{{{ graph( 500, 500, -11, 13, -11, 13,2*x^2+-5*x+-3) }}}


and we can see that the roots are {{{x=3}}} and {{{x=-1/2}}}. This verifies our answer