15=7t+4.9t^2

Solution for 15=7t+4.9t^2 equation:

15=7t+4.9t^2

We move all terms to the left:

15-(7t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-7t+15=0

a = -4.9; b = -7; c = +15;
Δ = b2-4ac
Δ = -72-4·(-4.9)·15
Δ = 343
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:

$\sqrt{\Delta}=\sqrt{343}=\sqrt{49*7}=\sqrt{49}*\sqrt{7}=7\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7\sqrt{7}}{2*-4.9}=\frac{7-7\sqrt{7}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7\sqrt{7}}{2*-4.9}=\frac{7+7\sqrt{7}}{-9.8}
$