4.9t^2+340t-1530=0

Solution for 4.9t^2+340t-1530=0 equation:

4.9t^2+340t-1530=0

a = 4.9; b = 340; c = -1530;
Δ = b2-4ac
Δ = 3402-4·4.9·(-1530)
Δ = 145588
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:

$\sqrt{\Delta}=\sqrt{145588}=\sqrt{4*36397}=\sqrt{4}*\sqrt{36397}=2\sqrt{36397}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(340)-2\sqrt{36397}}{2*4.9}=\frac{-340-2\sqrt{36397}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(340)+2\sqrt{36397}}{2*4.9}=\frac{-340+2\sqrt{36397}}{9.8}
$