5=-16t^2+28t+3

Solution for 5=-16t^2+28t+3 equation:

5=-16t^2+28t+3

We move all terms to the left:

5-(-16t^2+28t+3)=0

We get rid of parentheses

16t^2-28t-3+5=0

We add all the numbers together, and all the variables

16t^2-28t+2=0

a = 16; b = -28; c = +2;
Δ = b2-4ac
Δ = -282-4·16·2
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{41}}{2*16}=\frac{28-4\sqrt{41}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{41}}{2*16}=\frac{28+4\sqrt{41}}{32}
$