20=-16t^2+88+0

Solution for 20=-16t^2+88+0 equation:

20=-16t^2+88+0

We move all terms to the left:

20-(-16t^2+88+0)=0

We get rid of parentheses

16t^2-88-0+20=0

We add all the numbers together, and all the variables

16t^2-68=0

a = 16; b = 0; c = -68;
Δ = b2-4ac
Δ = 02-4·16·(-68)
Δ = 4352
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:

$\sqrt{\Delta}=\sqrt{4352}=\sqrt{256*17}=\sqrt{256}*\sqrt{17}=16\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{17}}{2*16}=\frac{0-16\sqrt{17}}{32}
=-\frac{16\sqrt{17}}{32}
=-\frac{\sqrt{17}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{17}}{2*16}=\frac{0+16\sqrt{17}}{32}
=\frac{16\sqrt{17}}{32}
=\frac{\sqrt{17}}{2}
$