0=-16t^2+148t

Solution for 0=-16t^2+148t equation:

0=-16t^2+148t

We move all terms to the left:

0-(-16t^2+148t)=0

We add all the numbers together, and all the variables

-(-16t^2+148t)=0

We get rid of parentheses

16t^2-148t=0

a = 16; b = -148; c = 0;
Δ = b2-4ac
Δ = -1482-4·16·0
Δ = 21904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{21904}=148$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-148)-148}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-148)+148}{2*16}=\frac{296}{32}
=9+1/4
$