Question 552892


{{{t^2+4t=1}}} Start with the given equation.



{{{t^2+4t-1=0}}} Get every term to the left side.



Notice that the quadratic {{{t^2+4t-1}}} is in the form of {{{At^2+Bt+C}}} where {{{A=1}}}, {{{B=4}}}, and {{{C=-1}}}



Let's use the quadratic formula to solve for "t":



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



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



{{{t = (-4 +- sqrt( 16-4(1)(-1) ))/(2(1))}}} Square {{{4}}} to get {{{16}}}. 



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



{{{t = (-4 +- sqrt( 16+4 ))/(2(1))}}} Rewrite {{{sqrt(16--4)}}} as {{{sqrt(16+4)}}}



{{{t = (-4 +- sqrt( 20 ))/(2(1))}}} Add {{{16}}} to {{{4}}} to get {{{20}}}



{{{t = (-4 +- sqrt( 20 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{t = (-4 +- 2*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>)  



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



{{{t = -2 +- sqrt(5)}}} Reduce.  



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



So the solutions are {{{t = -2+sqrt(5)}}} or {{{t = -2-sqrt(5)}}} 



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