Question 419764
equation is:


y^2 + 6y - 1 = 0


standard form of a quadratic equation is:


ax^2 + bx + c = 0


a is the coefficient of the x^2 term.
b is the coefficient of the x term.
c is the constant.


in your equation you have:


a = 1
b = 6
c = -1


the quadratic formula is:


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


you replace the a's and the b's and the c's with their respective values to get:


{{{x = (-6 +- sqrt(6^2-4*1*-1))/(2*1)}}}


the * is the multiplication symbol.


in the formula you see it looks like a little square.


I typed a *.   The formula generator program converted it to a little square.


you would simplify this expression to be shown as follows:


{{{x = (-6 +- sqrt(40))/2}}}


it can also be simplified further to be shown as:


{{{x = (-3) +- sqrt(40)/2}}}


it can even be simplified further to be shown as:


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


{{{sqrt(40)/2}}} is the same thing as {{{sqrt(10)}}} because {{{sqrt(40)}}} equals {{{sqrt(2*2*10)/2}}} which equals {{{2*sqrt(10)/2}}} which equals {{{sqrt(10)}}}.


since they only told you to leave the answer in radical form, then i suspect you're ok with any of the answers that still have a radical in them.


that's between you and your teacher as to how he/she wants to see the answer.


a graph of your equation looks like this:


{{{graph(600,600,-10,10,-20,20,x^2 + 6x - 1)}}}


you can see that the graph crosses the x-axis somewhere around x = -6.16 and x = .16 which is where the equation of {{{x = (-3) +- sqrt(10)}}} indicates.


note that sqrt(10) is somewhere around 3.16...