0=-16t^2+72t-70

Solution for 0=-16t^2+72t-70 equation:

0=-16t^2+72t-70

We move all terms to the left:

0-(-16t^2+72t-70)=0

We add all the numbers together, and all the variables

-(-16t^2+72t-70)=0

We get rid of parentheses

16t^2-72t+70=0

a = 16; b = -72; c = +70;
Δ = b2-4ac
Δ = -722-4·16·70
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-8\sqrt{11}}{2*16}=\frac{72-8\sqrt{11}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+8\sqrt{11}}{2*16}=\frac{72+8\sqrt{11}}{32}
$