(4.9t^2)+(2.25t)=0

Solution for (4.9t^2)+(2.25t)=0 equation:

(4.9t^2)+(2.25t)=0

We add all the numbers together, and all the variables

(4.9t^2)+(+2.25t)=0

We get rid of parentheses

4.9t^2+2.25t=0

a = 4.9; b = 2.25; c = 0;
Δ = b2-4ac
Δ = 2.252-4·4.9·0
Δ = 5.0625
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{-(2.25)-\sqrt{5.0625}}{2*4.9}=\frac{-2.25-\sqrt{5.0625}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.25)+\sqrt{5.0625}}{2*4.9}=\frac{-2.25+\sqrt{5.0625}}{9.8}
$