25-3=-16t^2+96t

Solution for 25-3=-16t^2+96t equation:

25-3=-16t^2+96t

We move all terms to the left:

25-3-(-16t^2+96t)=0

We add all the numbers together, and all the variables

-(-16t^2+96t)+22=0

We get rid of parentheses

16t^2-96t+22=0

a = 16; b = -96; c = +22;
Δ = b2-4ac
Δ = -962-4·16·22
Δ = 7808
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:

$\sqrt{\Delta}=\sqrt{7808}=\sqrt{64*122}=\sqrt{64}*\sqrt{122}=8\sqrt{122}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-8\sqrt{122}}{2*16}=\frac{96-8\sqrt{122}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+8\sqrt{122}}{2*16}=\frac{96+8\sqrt{122}}{32}
$