0=-4t^2+3t+1

Solution for 0=-4t^2+3t+1 equation:

0=-4t^2+3t+1

We move all terms to the left:

0-(-4t^2+3t+1)=0

We add all the numbers together, and all the variables

-(-4t^2+3t+1)=0

We get rid of parentheses

4t^2-3t-1=0

a = 4; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·4·(-1)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*4}=\frac{-2}{8}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*4}=\frac{8}{8}
=1
$