3t^2+18t+19=0

Solution for 3t^2+18t+19=0 equation:

3t^2+18t+19=0

a = 3; b = 18; c = +19;
Δ = b2-4ac
Δ = 182-4·3·19
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-4\sqrt{6}}{2*3}=\frac{-18-4\sqrt{6}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+4\sqrt{6}}{2*3}=\frac{-18+4\sqrt{6}}{6}
$