0=-t^2+4t+12

Solution for 0=-t^2+4t+12 equation:

0=-t^2+4t+12

We move all terms to the left:

0-(-t^2+4t+12)=0

We add all the numbers together, and all the variables

-(-t^2+4t+12)=0

We get rid of parentheses

t^2-4t-12=0

a = 1; b = -4; c = -12;
Δ = b2-4ac
Δ = -42-4·1·(-12)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*1}=\frac{-4}{2}
=-2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*1}=\frac{12}{2}
=6
$