10=19t+0.7t^2

Solution for 10=19t+0.7t^2 equation:

10=19t+0.7t^2

We move all terms to the left:

10-(19t+0.7t^2)=0

We get rid of parentheses

-0.7t^2-19t+10=0

a = -0.7; b = -19; c = +10;
Δ = b2-4ac
Δ = -192-4·(-0.7)·10
Δ = 389
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)-\sqrt{389}}{2*-0.7}=\frac{19-\sqrt{389}}{-1.4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{389}}{2*-0.7}=\frac{19+\sqrt{389}}{-1.4}
$