t=-16t^2+29t-12

Solution for t=-16t^2+29t-12 equation:

t=-16t^2+29t-12

We move all terms to the left:

t-(-16t^2+29t-12)=0

We get rid of parentheses

16t^2-29t+t+12=0

We add all the numbers together, and all the variables

16t^2-28t+12=0

a = 16; b = -28; c = +12;
Δ = b2-4ac
Δ = -282-4·16·12
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4}{2*16}=\frac{24}{32}
=3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*16}=\frac{32}{32}
=1
$