Question 89352
{{{x^2+10=19}}}


{{{x^2+10-19=0}}} Subtract 19 from both sides


Now 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+10*x-19=0}}} ( notice {{{a=1}}}, {{{b=10}}}, and {{{c=-19}}})


{{{x = (-10 +- sqrt( (10)^2-4*1*-19 ))/(2*1)}}} Plug in a=1, b=10, and c=-19




{{{x = (-10 +- sqrt( 100-4*1*-19 ))/(2*1)}}} Square 10 to get 100  




{{{x = (-10 +- sqrt( 100+76 ))/(2*1)}}} Multiply {{{-4*-19*1}}} to get {{{76}}}




{{{x = (-10 +- sqrt( 176 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-10 +- 4*sqrt(11))/(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 = (-10 +- 4*sqrt(11))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-10 + 4*sqrt(11))/2}}} or {{{x = (-10 - 4*sqrt(11))/2}}}



which split up to



{{{x=-10/2+4*sqrt(11)/2}}} or {{{x=-10/2-4*sqrt(11)/2}}}



and simplify to



{{{x=-5+2*sqrt(11)}}} or {{{x=-5-2*sqrt(11)}}}



Which approximate to


{{{x=1.6332495807108}}} or {{{x=-11.6332495807108}}}



So our solutions are:

{{{x=1.6332495807108}}} or {{{x=-11.6332495807108}}}


Notice when we graph {{{x^2+10*x-19}}}, we get:


{{{ graph( 500, 500, -21.6332495807108, 11.6332495807108, -21.6332495807108, 11.6332495807108,1*x^2+10*x+-19) }}}


when we use the root finder feature on a calculator, we find that {{{x=1.6332495807108}}} and {{{x=-11.6332495807108}}}.So this verifies our answer