0=4.9t^2-12t-40

Solution for 0=4.9t^2-12t-40 equation:

0=4.9t^2-12t-40

We move all terms to the left:

0-(4.9t^2-12t-40)=0

We add all the numbers together, and all the variables

-(4.9t^2-12t-40)=0

We get rid of parentheses

-4.9t^2+12t+40=0

a = -4.9; b = 12; c = +40;
Δ = b2-4ac
Δ = 122-4·(-4.9)·40
Δ = 928
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:

$\sqrt{\Delta}=\sqrt{928}=\sqrt{16*58}=\sqrt{16}*\sqrt{58}=4\sqrt{58}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{58}}{2*-4.9}=\frac{-12-4\sqrt{58}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{58}}{2*-4.9}=\frac{-12+4\sqrt{58}}{-9.8}
$