4.9t^2-20t+12=0

Solution for 4.9t^2-20t+12=0 equation:

4.9t^2-20t+12=0

a = 4.9; b = -20; c = +12;
Δ = b2-4ac
Δ = -202-4·4.9·12
Δ = 164.8
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{-(-20)-\sqrt{164.8}}{2*4.9}=\frac{20-\sqrt{164.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{164.8}}{2*4.9}=\frac{20+\sqrt{164.8}}{9.8}
$