-9t^2-3t+6=0

Solution for -9t^2-3t+6=0 equation:

-9t^2-3t+6=0

a = -9; b = -3; c = +6;
Δ = b2-4ac
Δ = -32-4·(-9)·6
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-15}{2*-9}=\frac{-12}{-18}
=2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+15}{2*-9}=\frac{18}{-18}
=-1
$