6t^2+13t-19=0

Solution for 6t^2+13t-19=0 equation:

6t^2+13t-19=0

a = 6; b = 13; c = -19;
Δ = b2-4ac
Δ = 132-4·6·(-19)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-25}{2*6}=\frac{-38}{12}
=-3+1/6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+25}{2*6}=\frac{12}{12}
=1
$