-5=15t-4.9t^2

Solution for -5=15t-4.9t^2 equation:

-5=15t-4.9t^2

We move all terms to the left:

-5-(15t-4.9t^2)=0

We get rid of parentheses

4.9t^2-15t-5=0

a = 4.9; b = -15; c = -5;
Δ = b2-4ac
Δ = -152-4·4.9·(-5)
Δ = 323
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{323}}{2*4.9}=\frac{15-\sqrt{323}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{323}}{2*4.9}=\frac{15+\sqrt{323}}{9.8}
$