0=4.9t^2+15t-40

Solution for 0=4.9t^2+15t-40 equation:

0=4.9t^2+15t-40

We move all terms to the left:

0-(4.9t^2+15t-40)=0

We add all the numbers together, and all the variables

-(4.9t^2+15t-40)=0

We get rid of parentheses

-4.9t^2-15t+40=0

a = -4.9; b = -15; c = +40;
Δ = b2-4ac
Δ = -152-4·(-4.9)·40
Δ = 1009
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{-(-15)-\sqrt{1009}}{2*-4.9}=\frac{15-\sqrt{1009}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{1009}}{2*-4.9}=\frac{15+\sqrt{1009}}{-9.8}
$