50=0.8t^2

Solution for 50=0.8t^2 equation:

50=0.8t^2

We move all terms to the left:

50-(0.8t^2)=0

We get rid of parentheses

-0.8t^2+50=0

a = -0.8; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-0.8)·50
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*-0.8}=\frac{0-4\sqrt{10}}{-1.6}
=-\frac{4\sqrt{10}}{-1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*-0.8}=\frac{0+4\sqrt{10}}{-1.6}
=\frac{4\sqrt{10}}{-1.6}
$