4.9t^2+25t-20=0

Solution for 4.9t^2+25t-20=0 equation:

4.9t^2+25t-20=0

a = 4.9; b = 25; c = -20;
Δ = b2-4ac
Δ = 252-4·4.9·(-20)
Δ = 1017
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:

$\sqrt{\Delta}=\sqrt{1017}=\sqrt{9*113}=\sqrt{9}*\sqrt{113}=3\sqrt{113}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-3\sqrt{113}}{2*4.9}=\frac{-25-3\sqrt{113}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+3\sqrt{113}}{2*4.9}=\frac{-25+3\sqrt{113}}{9.8}
$