4.9t^2+20t-300=0

Solution for 4.9t^2+20t-300=0 equation:

4.9t^2+20t-300=0

a = 4.9; b = 20; c = -300;
Δ = b2-4ac
Δ = 202-4·4.9·(-300)
Δ = 6280
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:

$\sqrt{\Delta}=\sqrt{6280}=\sqrt{4*1570}=\sqrt{4}*\sqrt{1570}=2\sqrt{1570}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{1570}}{2*4.9}=\frac{-20-2\sqrt{1570}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{1570}}{2*4.9}=\frac{-20+2\sqrt{1570}}{9.8}
$