15=-4.9t^2+21.46t

Solution for 15=-4.9t^2+21.46t equation:

15=-4.9t^2+21.46t

We move all terms to the left:

15-(-4.9t^2+21.46t)=0

We get rid of parentheses

4.9t^2-21.46t+15=0

a = 4.9; b = -21.46; c = +15;
Δ = b2-4ac
Δ = -21.462-4·4.9·15
Δ = 166.5316
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{-(-21.46)-\sqrt{166.5316}}{2*4.9}=\frac{21.46-\sqrt{166.5316}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21.46)+\sqrt{166.5316}}{2*4.9}=\frac{21.46+\sqrt{166.5316}}{9.8}
$