4(9-12t-4t^2)=0

Solution for 4(9-12t-4t^2)=0 equation:

4(9-12t-4t^2)=0

We multiply parentheses

-16t^2-48t+36=0

a = -16; b = -48; c = +36;
Δ = b2-4ac
Δ = -482-4·(-16)·36
Δ = 4608
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:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-48\sqrt{2}}{2*-16}=\frac{48-48\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48\sqrt{2}}{2*-16}=\frac{48+48\sqrt{2}}{-32}
$