4.9t^2-25t-50=0

Solution for 4.9t^2-25t-50=0 equation:

4.9t^2-25t-50=0

a = 4.9; b = -25; c = -50;
Δ = b2-4ac
Δ = -252-4·4.9·(-50)
Δ = 1605
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{-(-25)-\sqrt{1605}}{2*4.9}=\frac{25-\sqrt{1605}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{1605}}{2*4.9}=\frac{25+\sqrt{1605}}{9.8}
$