4=-16t^2+24t+3

Solution for 4=-16t^2+24t+3 equation:

4=-16t^2+24t+3

We move all terms to the left:

4-(-16t^2+24t+3)=0

We get rid of parentheses

16t^2-24t-3+4=0

We add all the numbers together, and all the variables

16t^2-24t+1=0

a = 16; b = -24; c = +1;
Δ = b2-4ac
Δ = -242-4·16·1
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-16\sqrt{2}}{2*16}=\frac{24-16\sqrt{2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+16\sqrt{2}}{2*16}=\frac{24+16\sqrt{2}}{32}
$