Question 103658
Solve by completing the square. 
:
2x^2 – 6x – 3 = 0
:
When completing the square, we need the coefficient of x^2 to be; divide eq by 2:
x^2 - 3x - {{{3/2}}} = 0
:
x^2 - 3x + ___ = {{{3/2}}}
:
Find the value that will make it a perfect square, divide the coefficient of x by 2 and square it; add to both sides:
{{{(3/2)^2}}} = {{{9/4}}}
we have:
x^2 - 3x + {{{9/4}}} = {{{3/2}}} + {{{9/4}}}
:
x^2 - 3x + {{{9/4}}} = {{{6/4}}} + {{{9/4}}}; use the common denominator on the right
:
x^2 - 3x + {{{9/4}}} = +{{{15/4}}}
:
(x - {{{3/2}}})^2 = {{{15/4}}}
:
x - {{{3/2}}} = +/- {{{sqrt(15/4)}}}; square root of both sides
:
x - {{{3/2}}} = +/- {{{(1/2)sqrt(15)}}}; extract the square root of 1/4
:
x = {{{3/2}}} +/- {{{(1/2)sqrt(15)}}}; Add {{{3/2}}} to both sides
Same as:
x = {{{(3 +- sqrt(15))/2}}}; 
Two solutions:
x = {{{(3 + sqrt(15))/2}}}
and
x = {{{(3 - sqrt(15))/2}}}