4.9t^2-23.8t+12=0

Solution for 4.9t^2-23.8t+12=0 equation:

4.9t^2-23.8t+12=0

a = 4.9; b = -23.8; c = +12;
Δ = b2-4ac
Δ = -23.82-4·4.9·12
Δ = 331.24
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{-(-23.8)-\sqrt{331.24}}{2*4.9}=\frac{23.8-\sqrt{331.24}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23.8)+\sqrt{331.24}}{2*4.9}=\frac{23.8+\sqrt{331.24}}{9.8}
$