Question 112883
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-2=0}}} ( notice {{{a=2}}}, {{{b=-3}}}, and {{{c=-2}}})





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




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




{{{x = (3 +- sqrt( 9-4*2*-2 ))/(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+16 ))/(2*2)}}} Multiply {{{-4*-2*2}}} to get {{{16}}}




{{{x = (3 +- sqrt( 25 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (3 +- 5)/(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 +- 5)/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{x = (3 + 5)/4}}} or {{{x = (3 - 5)/4}}}


Lets look at the first part:


{{{x=(3 + 5)/4}}}


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

{{{x=2}}} Divide


So one answer is

{{{x=2}}}




Now lets look at the second part:


{{{x=(3 - 5)/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=2}}} or {{{x=-1/2}}}


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


{{{ graph( 500, 500, -11, 12, -11, 12,2*x^2+-3*x+-2) }}}


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