15=4.9t^2

Solution for 15=4.9t^2 equation:

15=4.9t^2

We move all terms to the left:

15-(4.9t^2)=0

We get rid of parentheses

-4.9t^2+15=0

a = -4.9; b = 0; c = +15;
Δ = b2-4ac
Δ = 02-4·(-4.9)·15
Δ = 294
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:

$\sqrt{\Delta}=\sqrt{294}=\sqrt{49*6}=\sqrt{49}*\sqrt{6}=7\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-7\sqrt{6}}{2*-4.9}=\frac{0-7\sqrt{6}}{-9.8}
=-\frac{7\sqrt{6}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+7\sqrt{6}}{2*-4.9}=\frac{0+7\sqrt{6}}{-9.8}
=\frac{7\sqrt{6}}{-9.8}
$