Question 299452
{{{t^2+10t+26=0}}} Start with the given equation.



Notice that the quadratic {{{t^2+10t+26}}} is in the form of {{{At^2+Bt+C}}} where {{{A=1}}}, {{{B=10}}}, and {{{C=26}}}



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



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



{{{t = (-(10) +- sqrt( (10)^2-4(1)(26) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=10}}}, and {{{C=26}}}



{{{t = (-10 +- sqrt( 100-4(1)(26) ))/(2(1))}}} Square {{{10}}} to get {{{100}}}. 



{{{t = (-10 +- sqrt( 100-104 ))/(2(1))}}} Multiply {{{4(1)(26)}}} to get {{{104}}}



{{{t = (-10 +- sqrt( -4 ))/(2(1))}}} Subtract {{{104}}} from {{{100}}} to get {{{-4}}}



{{{t = (-10 +- sqrt( -4 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{t = (-10 +- 2*i)/(2)}}} Take the square root of {{{-4}}} to get {{{2*i}}}. 



{{{t = (-10 + 2*i)/(2)}}} or {{{t = (-10 - 2*i)/(2)}}} Break up the expression. 



{{{t = (-10)/(2) + (2*i)/(2)}}} or {{{t =  (-10)/(2) - (2*i)/(2)}}} Break up the fraction for each case. 



{{{t = -5+1*i}}} or {{{t = -5-i}}} Reduce. 



So the solutions are {{{t = -5+i}}} or {{{t = -5-i}}}