4.9t^2+130t-2100=0

Solution for 4.9t^2+130t-2100=0 equation:

4.9t^2+130t-2100=0

a = 4.9; b = 130; c = -2100;
Δ = b2-4ac
Δ = 1302-4·4.9·(-2100)
Δ = 58060
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:

$\sqrt{\Delta}=\sqrt{58060}=\sqrt{4*14515}=\sqrt{4}*\sqrt{14515}=2\sqrt{14515}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(130)-2\sqrt{14515}}{2*4.9}=\frac{-130-2\sqrt{14515}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(130)+2\sqrt{14515}}{2*4.9}=\frac{-130+2\sqrt{14515}}{9.8}
$