2.5+70t-16t^2=0

Solution for 2.5+70t-16t^2=0 equation:

2.5+70t-16t^2=0

a = -16; b = 70; c = +2.5;
Δ = b2-4ac
Δ = 702-4·(-16)·2.5
Δ = 5060
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:

$\sqrt{\Delta}=\sqrt{5060}=\sqrt{4*1265}=\sqrt{4}*\sqrt{1265}=2\sqrt{1265}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{1265}}{2*-16}=\frac{-70-2\sqrt{1265}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{1265}}{2*-16}=\frac{-70+2\sqrt{1265}}{-32}
$