0=16t^2+45t-3

Solution for 0=16t^2+45t-3 equation:

0=16t^2+45t-3

We move all terms to the left:

0-(16t^2+45t-3)=0

We add all the numbers together, and all the variables

-(16t^2+45t-3)=0

We get rid of parentheses

-16t^2-45t+3=0

a = -16; b = -45; c = +3;
Δ = b2-4ac
Δ = -452-4·(-16)·3
Δ = 2217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-\sqrt{2217}}{2*-16}=\frac{45-\sqrt{2217}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{2217}}{2*-16}=\frac{45+\sqrt{2217}}{-32}
$