19t^2+6t-13=0

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

19t^2+6t-13=0

a = 19; b = 6; c = -13;
Δ = b2-4ac
Δ = 62-4·19·(-13)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-32}{2*19}=\frac{-38}{38}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+32}{2*19}=\frac{26}{38}
=13/19
$