Question 61998
2t^2 – 19 – 6t = 0

2t^2 -6t - 19 = 0 --->(1)
 
It is the quadratic equation of the form at^2+ bt + c = 0 --->(2)

When we compare the two equations we get,

a =2, b  = -6, c = -19

We shall substitue in the quadratic formula, we get

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-(-6) +- sqrt( (-6)^2-4*2*-19 ))/(2*2) }}} 
{{{x = (+6 +- sqrt( 36+152 ))/(4) }}} 
{{{x = (+6 +- sqrt( 188 ))/(4) }}} 
{{{x = (+6 +- 2sqrt( 47 ))/(4) }}} 
{{{x = (+3 +- sqrt( 47 ))/(2) }}}