-3t^2+18t-11=0

Solution for -3t^2+18t-11=0 equation:

-3t^2+18t-11=0

a = -3; b = 18; c = -11;
Δ = b2-4ac
Δ = 182-4·(-3)·(-11)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-8\sqrt{3}}{2*-3}=\frac{-18-8\sqrt{3}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+8\sqrt{3}}{2*-3}=\frac{-18+8\sqrt{3}}{-6}
$