292=-16t^2+145t+3

Solution for 292=-16t^2+145t+3 equation:

292=-16t^2+145t+3

We move all terms to the left:

292-(-16t^2+145t+3)=0

We get rid of parentheses

16t^2-145t-3+292=0

We add all the numbers together, and all the variables

16t^2-145t+289=0

a = 16; b = -145; c = +289;
Δ = b2-4ac
Δ = -1452-4·16·289
Δ = 2529
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:

$\sqrt{\Delta}=\sqrt{2529}=\sqrt{9*281}=\sqrt{9}*\sqrt{281}=3\sqrt{281}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-145)-3\sqrt{281}}{2*16}=\frac{145-3\sqrt{281}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-145)+3\sqrt{281}}{2*16}=\frac{145+3\sqrt{281}}{32}
$