0=25+5.3t-4.9t^2

Solution for 0=25+5.3t-4.9t^2 equation:

0=25+5.3t-4.9t^2

We move all terms to the left:

0-(25+5.3t-4.9t^2)=0

We add all the numbers together, and all the variables

-(25+5.3t-4.9t^2)=0

We get rid of parentheses

4.9t^2-5.3t-25=0

a = 4.9; b = -5.3; c = -25;
Δ = b2-4ac
Δ = -5.32-4·4.9·(-25)
Δ = 518.09
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5.3)-\sqrt{518.09}}{2*4.9}=\frac{5.3-\sqrt{518.09}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5.3)+\sqrt{518.09}}{2*4.9}=\frac{5.3+\sqrt{518.09}}{9.8}
$