4.9t^2-21t-33=0

Solution for 4.9t^2-21t-33=0 equation:

4.9t^2-21t-33=0

a = 4.9; b = -21; c = -33;
Δ = b2-4ac
Δ = -212-4·4.9·(-33)
Δ = 1087.8
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{-(-21)-\sqrt{1087.8}}{2*4.9}=\frac{21-\sqrt{1087.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{1087.8}}{2*4.9}=\frac{21+\sqrt{1087.8}}{9.8}
$