12=19t-5t^2

Solution for 12=19t-5t^2 equation:

12=19t-5t^2

We move all terms to the left:

12-(19t-5t^2)=0

We get rid of parentheses

5t^2-19t+12=0

a = 5; b = -19; c = +12;
Δ = b2-4ac
Δ = -192-4·5·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-11}{2*5}=\frac{8}{10}
=4/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+11}{2*5}=\frac{30}{10}
=3
$