Question 197756


{{{x^2+8x-4=0}}} Start with the given equation.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=8}}}, 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 = (-(8) +- sqrt( (8)^2-4(1)(-4) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=8}}}, and {{{C=-4}}}



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



{{{x = (-8 +- sqrt( 64--16 ))/(2(1))}}} Multiply {{{4(1)(-4)}}} to get {{{-16}}}



{{{x = (-8 +- sqrt( 64+16 ))/(2(1))}}} Rewrite {{{sqrt(64--16)}}} as {{{sqrt(64+16)}}}



{{{x = (-8 +- sqrt( 80 ))/(2(1))}}} Add {{{64}}} to {{{16}}} to get {{{80}}}



{{{x = (-8 +- sqrt( 80 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-8 +- 4*sqrt(5))/(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 = (-8)/(2) +- (4*sqrt(5))/(2)}}} Break up the fraction.  



{{{x = -4 +- 2*sqrt(5)}}} Reduce.  



{{{x = -4+2*sqrt(5)}}} or {{{x = -4-2*sqrt(5)}}} Break up the expression.  



So the solutions are {{{x = -4+2*sqrt(5)}}} or {{{x = -4-2*sqrt(5)}}} 



which approximate to {{{x=0.472}}} or {{{x=-8.472}}}