Question 112101
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 {{{x^2+7*x+10=0}}} ( notice {{{a=1}}}, {{{b=7}}}, and {{{c=10}}})





{{{x = (-7 +- sqrt( (7)^2-4*1*10 ))/(2*1)}}} Plug in a=1, b=7, and c=10




{{{x = (-7 +- sqrt( 49-4*1*10 ))/(2*1)}}} Square 7 to get 49  




{{{x = (-7 +- sqrt( 49+-40 ))/(2*1)}}} Multiply {{{-4*10*1}}} to get {{{-40}}}




{{{x = (-7 +- sqrt( 9 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-7 +- 3)/(2*1)}}} 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 = (-7 +- 3)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-7 + 3)/2}}} or {{{x = (-7 - 3)/2}}}


Lets look at the first part:


{{{x=(-7 + 3)/2}}}


{{{x=-4/2}}} Add the terms in the numerator

{{{x=-2}}} Divide


So one answer is

{{{x=-2}}}




Now lets look at the second part:


{{{x=(-7 - 3)/2}}}


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

{{{x=-5}}} Divide


So another answer is

{{{x=-5}}}


So our solutions are:

{{{x=-2}}} or {{{x=-5}}}


Notice when we graph {{{x^2+7*x+10}}}, we get:


{{{ graph( 500, 500, -15, 8, -15, 8,1*x^2+7*x+10) }}}


and we can see that the roots are {{{x=-2}}} and {{{x=-5}}}. This verifies our answer