4.9t^2-24t+12=0

Solution for 4.9t^2-24t+12=0 equation:

4.9t^2-24t+12=0

a = 4.9; b = -24; c = +12;
Δ = b2-4ac
Δ = -242-4·4.9·12
Δ = 340.8
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{-(-24)-\sqrt{340.8}}{2*4.9}=\frac{24-\sqrt{340.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+\sqrt{340.8}}{2*4.9}=\frac{24+\sqrt{340.8}}{9.8}
$