Question 89288
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+8*x-20=0}}} ( notice {{{a=1}}}, {{{b=8}}}, and {{{c=-20}}})


{{{x = (-8 +- sqrt( (8)^2-4*1*-20 ))/(2*1)}}} Plug in a=1, b=8, and c=-20




{{{x = (-8 +- sqrt( 64-4*1*-20 ))/(2*1)}}} Square 8 to get 64




{{{x = (-8 +- sqrt( 64+80 ))/(2*1)}}} Multiply {{{-4*-20*1}}} to get {{{80}}}




{{{x = (-8 +- sqrt( 144 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-8 +- 12)/(2*1)}}} Simplify the square root




{{{x = (-8 +- 12)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-8 + 12)/2}}} or {{{x = (-8 - 12)/2}}}


Lets look at the first part:


{{{x=(-8 + 12)/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=(-8 - 12)/2}}}


{{{x=-20/2}}} Subtract the terms in the numerator

{{{x=-10}}} Divide


So another answer is

{{{x=-10}}}


So our solutions are:

{{{x=2}}} or {{{x=-10}}}


Notice when we graph {{{x^2+8*x-20}}}, we get:


{{{ graph( 500, 500, -20, 12, -20, 12,1*x^2+8*x+-20) }}}


and we can see that the roots are {{{x=2}}} and {{{x=-10}}}. This verifies our answer