-3t^2+18t-21=0

Solution for -3t^2+18t-21=0 equation:

-3t^2+18t-21=0

a = -3; b = 18; c = -21;
Δ = b2-4ac
Δ = 182-4·(-3)·(-21)
Δ = 72
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:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{2}}{2*-3}=\frac{-18-6\sqrt{2}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{2}}{2*-3}=\frac{-18+6\sqrt{2}}{-6}
$