0=-4t^2+32t-t

Solution for 0=-4t^2+32t-t equation:

0=-4t^2+32t-t

We move all terms to the left:

0-(-4t^2+32t-t)=0

We add all the numbers together, and all the variables

-(-4t^2+32t-t)=0

We get rid of parentheses

4t^2-32t+t=0

We add all the numbers together, and all the variables

4t^2-31t=0

a = 4; b = -31; c = 0;
Δ = b2-4ac
Δ = -312-4·4·0
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-31}{2*4}=\frac{0}{8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+31}{2*4}=\frac{62}{8}
=7+3/4
$