Question 88264
Now let's use the quadratic formula to solve for x:



Starting with the general quadratic


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


the general form of the quadratic equation is:


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


So lets solve {{{x^2+8*x+15=0}}} (notice {{{a=1}}}, {{{b=8}}}, and {{{c=15}}})


{{{x = (-8 +- sqrt( (8)^2-4*1*15 ))/(2*1)}}} Plug in a=1, b=8, and c=15




{{{x = (-8 +- sqrt( 64-4*1*15 ))/(2*1)}}} Square 8 to get 64




{{{x = (-8 +- sqrt( 64+-60 ))/(2*1)}}} Multiply {{{-4*15*1}}} to get {{{-60}}}




{{{x = (-8 +- sqrt( 4 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-8 +- 2)/(2*1)}}} Simplify the square root




{{{x = (-8 +- 2)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-8 + 2)/2}}} or {{{x = (-8 - 2)/2}}}


Lets look at the first part:


{{{x=-6/2}}} Add the terms in the numerator

{{{x=-3}}} Divide


So one answer is

{{{x=-3}}}

Now lets look at the second part:


{{{x=-10/2}}} Subtract the terms in the numerator

{{{x=-5}}} Divide


So another answer is

{{{x=-5}}}


So our solutions are:

{{{x=-3}}} or {{{x=-5}}}


Notice when we graph {{{x^2+8*x+15}}} we get:


{{{ graph( 500, 500, -15, 7, -15, 7,1*x^2+8*x+15) }}}


and we can see that the roots are {{{x=-3}}} and {{{x=-5}}}. This verifies our answer