8=-16t^2+140

Solution for 8=-16t^2+140 equation:

8=-16t^2+140

We move all terms to the left:

8-(-16t^2+140)=0

We get rid of parentheses

16t^2-140+8=0

We add all the numbers together, and all the variables

16t^2-132=0

a = 16; b = 0; c = -132;
Δ = b2-4ac
Δ = 02-4·16·(-132)
Δ = 8448
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:

$\sqrt{\Delta}=\sqrt{8448}=\sqrt{256*33}=\sqrt{256}*\sqrt{33}=16\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{33}}{2*16}=\frac{0-16\sqrt{33}}{32}
=-\frac{16\sqrt{33}}{32}
=-\frac{\sqrt{33}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{33}}{2*16}=\frac{0+16\sqrt{33}}{32}
=\frac{16\sqrt{33}}{32}
=\frac{\sqrt{33}}{2}
$