15=t+4.9t^2

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

15=t+4.9t^2

We move all terms to the left:

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

We get rid of parentheses

-4.9t^2-t+15=0

We add all the numbers together, and all the variables

-4.9t^2-1t+15=0

a = -4.9; b = -1; c = +15;
Δ = b2-4ac
Δ = -12-4·(-4.9)·15
Δ = 295
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{-(-1)-\sqrt{295}}{2*-4.9}=\frac{1-\sqrt{295}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{295}}{2*-4.9}=\frac{1+\sqrt{295}}{-9.8}
$