-4.9t^2+36t+25=0

Solution for -4.9t^2+36t+25=0 equation:

-4.9t^2+36t+25=0

a = -4.9; b = 36; c = +25;
Δ = b2-4ac
Δ = 362-4·(-4.9)·25
Δ = 1786
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{-(36)-\sqrt{1786}}{2*-4.9}=\frac{-36-\sqrt{1786}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+\sqrt{1786}}{2*-4.9}=\frac{-36+\sqrt{1786}}{-9.8}
$