70=19t+5t^2

Solution for 70=19t+5t^2 equation:

70=19t+5t^2

We move all terms to the left:

70-(19t+5t^2)=0

We get rid of parentheses

-5t^2-19t+70=0

a = -5; b = -19; c = +70;
Δ = b2-4ac
Δ = -192-4·(-5)·70
Δ = 1761
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{1761}}{2*-5}=\frac{19-\sqrt{1761}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{1761}}{2*-5}=\frac{19+\sqrt{1761}}{-10}
$