4.9t2+5.8t-21=0

Solution for 4.9t2+5.8t-21=0 equation:

4.9t^2+5.8t-21=0

a = 4.9; b = 5.8; c = -21;
Δ = b2-4ac
Δ = 5.82-4·4.9·(-21)
Δ = 445.24
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{-(5.8)-\sqrt{445.24}}{2*4.9}=\frac{-5.8-\sqrt{445.24}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5.8)+\sqrt{445.24}}{2*4.9}=\frac{-5.8+\sqrt{445.24}}{9.8}
$