-4.9(t-13)(t+2)=0

Solution for -4.9(t-13)(t+2)=0 equation:

-4.9(t-13)(t+2)=0

We multiply parentheses ..

-4.9(+t^2+2t-13t-26)=0

We multiply parentheses

-4.9t^2-9.8t+63.7t+127.4=0

We add all the numbers together, and all the variables

-4.9t^2+53.9t+127.4=0

a = -4.9; b = 53.9; c = +127.4;
Δ = b2-4ac
Δ = 53.92-4·(-4.9)·127.4
Δ = 5402.25
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{-(53.9)-\sqrt{5402.25}}{2*-4.9}=\frac{-53.9-\sqrt{5402.25}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(53.9)+\sqrt{5402.25}}{2*-4.9}=\frac{-53.9+\sqrt{5402.25}}{-9.8}
$