-16(t^2-4)=-4

Solution for -16(t^2-4)=-4 equation:

-16(t^2-4)=-4

We move all terms to the left:

-16(t^2-4)-(-4)=0

We add all the numbers together, and all the variables

-16(t^2-4)+4=0

We multiply parentheses

-16t^2+64+4=0

We add all the numbers together, and all the variables

-16t^2+68=0

a = -16; b = 0; c = +68;
Δ = b2-4ac
Δ = 02-4·(-16)·68
Δ = 4352
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:

$\sqrt{\Delta}=\sqrt{4352}=\sqrt{256*17}=\sqrt{256}*\sqrt{17}=16\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{17}}{2*-16}=\frac{0-16\sqrt{17}}{-32}
=-\frac{16\sqrt{17}}{-32}
=-\frac{\sqrt{17}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{17}}{2*-16}=\frac{0+16\sqrt{17}}{-32}
=\frac{16\sqrt{17}}{-32}
=\frac{\sqrt{17}}{-2}
$