19=-8t^2+24t+3

Solution for 19=-8t^2+24t+3 equation:

19=-8t^2+24t+3

We move all terms to the left:

19-(-8t^2+24t+3)=0

We get rid of parentheses

8t^2-24t-3+19=0

We add all the numbers together, and all the variables

8t^2-24t+16=0

a = 8; b = -24; c = +16;
Δ = b2-4ac
Δ = -242-4·8·16
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{-(-24)-8}{2*8}=\frac{16}{16}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8}{2*8}=\frac{32}{16}
=2
$