-3t^2+12t+36=0

Solution for -3t^2+12t+36=0 equation:

-3t^2+12t+36=0

a = -3; b = 12; c = +36;
Δ = b2-4ac
Δ = 122-4·(-3)·36
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*-3}=\frac{-36}{-6}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*-3}=\frac{12}{-6}
=-2
$