12+64t-16t^2=40

Solution for 12+64t-16t^2=40 equation:

12+64t-16t^2=40

We move all terms to the left:

12+64t-16t^2-(40)=0

We add all the numbers together, and all the variables

-16t^2+64t-28=0

a = -16; b = 64; c = -28;
Δ = b2-4ac
Δ = 642-4·(-16)·(-28)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-48}{2*-16}=\frac{-112}{-32}
=3+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+48}{2*-16}=\frac{-16}{-32}
=1/2
$