3+70t-16t^2=0

Solution for 3+70t-16t^2=0 equation:

3+70t-16t^2=0

a = -16; b = 70; c = +3;
Δ = b2-4ac
Δ = 702-4·(-16)·3
Δ = 5092
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5092}=\sqrt{4*1273}=\sqrt{4}*\sqrt{1273}=2\sqrt{1273}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{1273}}{2*-16}=\frac{-70-2\sqrt{1273}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{1273}}{2*-16}=\frac{-70+2\sqrt{1273}}{-32}
$