-4(4t^2-8t+3)=0

Solution for -4(4t^2-8t+3)=0 equation:

-4(4t^2-8t+3)=0

We multiply parentheses

-16t^2+32t-12=0

a = -16; b = 32; c = -12;
Δ = b2-4ac
Δ = 322-4·(-16)·(-12)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16}{2*-16}=\frac{-48}{-32}
=1+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16}{2*-16}=\frac{-16}{-32}
=1/2
$