Question 292613
<pre><font size = 4 color = "indigo"><b>
When you get here:

{{{5x^2-3x-1=0}}}

Compare that to

{{{ax^2+bx+c=0}}}

and you can see that {{{a=5}}}, {{{""+b=-3}}} and {{{""+c=-1}}}

Then you write

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

Now substitute {{{(5)}}} for {{{a}}}, {{{(-3)}}} for {{{b}}}, and {{{(-1)}}} for c
in that equation, and when you do you get:

{{{x = (-(-3) +- sqrt( (-3)^2-4*(5)*(-1) ))/(2*(5)) }}}

Then you simplify that:

{{{x = (3 +- sqrt(9+20 ))/10 }}}

{{{x = (3 +- sqrt(29 ))/10 }}}

So you have two solutions.

Solution 1 has a + where the ± is

{{{x = (3 + sqrt(29))/10 }}}
Then use your calculator to get the approximation for solution 1:
{{{x = (3 + sqrt(29))/10= (3+5.385164807)/10 = 8.385164807/10 = 0.8385164807}}} 
 
Solution 2 has a - where the ± is

{{{x = (3 - sqrt(29))/10 }}}
Then use your calculator to get the approximation for solution 2:
{{{x = (3 - sqrt(29))/10= (3-5.385164807)/10 = -2.385164807/10 = -0.28385164807}}}

Edwin</pre>