-3t^2+12t-9=0

Solution for -3t^2+12t-9=0 equation:

-3t^2+12t-9=0

a = -3; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·(-3)·(-9)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6}{2*-3}=\frac{-18}{-6}
=+3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6}{2*-3}=\frac{-6}{-6}
=1
$