22=15t+4.9t^2

Solution for 22=15t+4.9t^2 equation:

22=15t+4.9t^2

We move all terms to the left:

22-(15t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-15t+22=0

a = -4.9; b = -15; c = +22;
Δ = b2-4ac
Δ = -152-4·(-4.9)·22
Δ = 656.2
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{656.2}}{2*-4.9}=\frac{15-\sqrt{656.2}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{656.2}}{2*-4.9}=\frac{15+\sqrt{656.2}}{-9.8}
$