-4.9t^2+49t+15=0

Solution for -4.9t^2+49t+15=0 equation:

-4.9t^2+49t+15=0

a = -4.9; b = 49; c = +15;
Δ = b2-4ac
Δ = 492-4·(-4.9)·15
Δ = 2695
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:

$\sqrt{\Delta}=\sqrt{2695}=\sqrt{49*55}=\sqrt{49}*\sqrt{55}=7\sqrt{55}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-7\sqrt{55}}{2*-4.9}=\frac{-49-7\sqrt{55}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+7\sqrt{55}}{2*-4.9}=\frac{-49+7\sqrt{55}}{-9.8}
$