Question 378722


{{{x^2+10x-4=0}}} Start with the given equation.



Notice that the quadratic {{{x^2+10x-4}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=10}}}, and {{{C=-4}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(10) +- sqrt( (10)^2-4(1)(-4) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=10}}}, and {{{C=-4}}}



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



{{{x = (-10 +- sqrt( 100--16 ))/(2(1))}}} Multiply {{{4(1)(-4)}}} to get {{{-16}}}



{{{x = (-10 +- sqrt( 100+16 ))/(2(1))}}} Rewrite {{{sqrt(100--16)}}} as {{{sqrt(100+16)}}}



{{{x = (-10 +- sqrt( 116 ))/(2(1))}}} Add {{{100}}} to {{{16}}} to get {{{116}}}



{{{x = (-10 +- sqrt( 116 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-10 +- 2*sqrt(29))/(2)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (-10)/(2) +- (2*sqrt(29))/(2)}}} Break up the fraction.  



{{{x = -5 +- sqrt(29)}}} Reduce.  



{{{x = -5+sqrt(29)}}} or {{{x = -5-sqrt(29)}}} Break up the expression.  



So the solutions are {{{x = -5+sqrt(29)}}} or {{{x = -5-sqrt(29)}}}



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Jim