0=-t^2+19.4t+12

Solution for 0=-t^2+19.4t+12 equation:

0=-t^2+19.4t+12

We move all terms to the left:

0-(-t^2+19.4t+12)=0

We add all the numbers together, and all the variables

-(-t^2+19.4t+12)=0

We get rid of parentheses

t^2-19.4t-12=0

a = 1; b = -19.4; c = -12;
Δ = b2-4ac
Δ = -19.42-4·1·(-12)
Δ = 424.36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19.4)-\sqrt{424.36}}{2*1}=\frac{19.4-\sqrt{424.36}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19.4)+\sqrt{424.36}}{2*1}=\frac{19.4+\sqrt{424.36}}{2}
$