23t-4.9t^2=0

Solution for 23t-4.9t^2=0 equation:

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