4.9t2+-23t+12=0

Solution for 4.9t2+-23t+12=0 equation:

4.9t^2+-23t+12=0

We add all the numbers together, and all the variables

4.9t^2-23t=0

a = 4.9; b = -23; c = 0;
Δ = b2-4ac
Δ = -232-4·4.9·0
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-23}{2*4.9}=\frac{0}{9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+23}{2*4.9}=\frac{46}{9.8}
=4+1.7894736842105/2.5789473684211
$